CIVL 433

Highway Engineering

Course Instructor
Dr. Ammar Alalkim Alzaabi

Lecture 7: Offsets for Vertical Curves

SSD and Crest Vertical Curve Design
Offsets for equal tangent Vertical Curves

• Now that we have learnt how to determine the elements of a

vertical curve design and are able to calculate vertical curve
stations and elevation, some additional properties of vertical
curves can now be formalized.

• Offsets are vertical distances from the initial tangent to the

curve. They are extremely important in vertical curve design
and construction.
Offsets for equal tangent Vertical Curves

G1 = initial roadway grade in % or m/m (initial L = length of the curve in stations or m measured in a
tangent grade). constant-elevation horizontal plane

G2 = final roadway (tangent) grade in % or m/m PVC = Point of the vertical ( the initial point of the curve)

Y = Offset at any distance x from the PVC in m PVI = Point of the vertical intersection ( intersection of
initial and final grades)
Ym = midcurve offset in m
PVT = point of vertical tangent, which is the final point of
Yf = offset at the end of the vertical curve in m the vertical curve ( the point where the curve returns to
the final grade or, the final tangent).
x = distance from the PVC in m
Offsets for equal tangent Vertical Curves

The properties of an equal-tangent parabola can be used to give

𝐴
Y = 200𝐿 𝑥 2 (7)

A = absolute value of the difference in grades ( lG1 – G2l )

expressed in %. 200 is used in the denominator instead of 2
because A is expressed in percent instead of m/m.

𝐴𝐿
Ym = 800
(8)

and

𝐴𝐿
Yf = (9)
200
Offsets for equal tangent Vertical Curves

This K-value can also be used to compute the high and low point
location of crest and sag vertical curves, respectively ( provided
the high or low point does not occur at the PVC or PVT).

If eq.6 is used to substitute a in eq. 2 (with L = KA), it can be

shown that setting dy/dx=0 in eq. 2 gives:

𝐿
K=𝐴 (10)

Where,

K = value that is the horizontal distance, in m, required to affect a

1% change in the slope of the vertical curve

L = length of the curve in m, and

A = absolute value of the difference in grades ( lG1 – G2l )

expressed as a percentage
Offsets for equal tangent Vertical Curves

Length of curve required to effect a 1% change in slope. Eq.1

gives a constant rate of change of slope and therefore it can be
shown that the horizontal distance required to change the slope
by 1% is

𝑥ℎ𝑙 = K x lG1l (11)

Where,

𝑥ℎ𝑙 = distance from the PVC to the high/low point in m, and other
terms are as defined previously

K- values have an important application in stopping sight distance

and vertical curve design which will be covered at later stage of
this lecture
Ex1: Vertical Curve Design with K-Values
A curve has initial and final grades +3% and -4%, respectively
and is 210m long. The PVC is at elevation 100m. Graph the
vertical curve elevations and the slope of the curve against the
length of the curve. Compute the K-value and use it to locate the
high point of the curve (distance from the PVC)

To find the slope at any point on the curve, take the

derivative of eq.1 to give us eq. 2.

𝑑𝑦
𝑑𝑥
= 2𝑎𝑥 + 𝑏 (2)

To solve this equation, we need to determine a and b.

𝐺2 −𝐺1 −0.04−(0.03)
Where, 𝑎= 2𝐿
= 2(210)
= −0.0001666

And 𝑏 = 𝐺1 = 0.03
Ex1: Vertical Curve Design with K-Values
𝐿 210
K=𝐴= = 30
3− −4
The circular points on the slope of the curve line correspond to changes
of 1% in grade, and these points occur at equal intervals of 30m.

This indicates that there should be a change in grade of 1% for every

30m of curve length ( measured in the horizontal plane).

Computing the K-value into eq.11 give us the high point value as follows:

𝑥ℎ𝑙 = K x lG1l = 30 x 3 = 90m

This indicates that the slope of the curve at 90m is zero.

The k-value gives the horizontal distance required to effect a 1% change

in the slope of the curve, and for this curve, the value is 30m.

Therefore, to go from an intial grade (G1) of 3% to a grade of 0% (the high

point), it takes a horizontal distance equal to K x 3, or 90m.
Ex1: Vertical Curve Design with K-Values

The K-value can also be solved by setting the derivative of eq.2 = 0 and
solving for 𝑥.

0 = 2𝑎𝑥 + 𝑏

Step 1 0=(2)(-0.000167)𝑥+0.03

0=-0.000333𝑥+0.03

Step 2 -0.000333𝑥 = -0.03

−0.03
Step 3 𝑥 = −0.000333

𝑥 = 90.09009
Ex1: Vertical Curve Design with K-Values
𝐿 210
K=𝐴= = 30
3− −4
Ex2: Vertical Curve Design Using Offsets
A vertical curve crosses a 1m diameter pipe at right angles. The pipe
is located at station 3 + 420 and its centerline is at elevation 333m.
The PVI of the vertical curve is at station 3+400 and elevation 335m.
The vertical curve is equal tangent, 180m long, and connects an
initial grade of +1.20% and a final grade of – 1.08%.

Using offsets, determine the depth, below the surface of the curve,
of the top of the pipe and determine the station of the highest point
of the curve.

The PVC is at station 3+310 ( 3+400minus 0 + 090, which is half of

the curve length), so the pipe is 110m (3+420 minus 3+310) from the
beginning of the curve (PVC). The elevation of the PVC will be the
elevation of the PVI minus the drop in grade over one-half the curve
length,

335 – (90m x 0.012m/m) = 333.92m

Ex2: Vertical Curve Design Using Offsets

335 – (90m x 0.012m/m) = 333.92m

Using this the elevation of the initial tangent above the pipe is

333.92 + (110m x 0.012 m/m) = 335.24m

Using eq. 7 to determine the offset above the pipe at 𝑥 = 110m ( the
distance of the pipe from the PVC), we have

𝐴
Y= 𝑥2
200𝐿

1.2− −1.08 2
Y= 110 = 0.77𝑚
200 180
Ex2: Vertical Curve Design Using Offsets

Thus the elevation of the curve above the pipe is 334.47m (335.24 – 0.77).
The elevation of the top of the pipe is 333.5m ( elevation of the centerline +
one half of the pipe’s diameter), so the pipe is 0.97m bellow the surface of
the curve (334.47 – 333.5).

Using eq. 10, we determine the location of the highest point on the curve,

𝐿 180
K=𝐴= 1.2− −1.08
= 78.95m

Using eq 11. the distance from the PVC to the highest point is

𝑥ℎ𝑙 = K x lG1l = 78.95 x 1.2 = 94.73m

Therefore the station of the highest point is at 3 + 404.73 ( 3+310 +

0+094.73)
Ex2: Vertical Curve Design Using Offsets

Again, determining the location of the highest point can also be solved by
equating the 1st derivative of eq.1 (eq.2) to zero and solving for 𝑥. Similar
to the calculation carried out in Ex1.

𝑑𝑦
= 2𝑎𝑥 + 𝑏 (2)
𝑑𝑥
Stopping Sight Distance (SSD)

Construction of a vertical curve is generally a costly operation requiring the

movement of significant amount of earthen material.

A main objective for highway designers is to minimize construction costs.

Thus, as engineers, we strive to make the vertical curve as short as safety
permits.

The level of safety is defined by the sufficient sight distance given to the
driver that would allow them to come to a complete stop, avoiding the
collision into objects obstructing their forward motion.

Adequate roadway drainage is also important but it is not covered in this

course.
Stopping Sight Distance and Crest Vertical
Curve Design

The length of the curve is the critical element in providing sufficient SSD
on a vertical curve.

The longer the curve, the more SSD is provided to the driver, but the
higher the cost as well.

Shorter curve lengths are cheaper to construct but may not provide
adequate SSD due to more rapid changes in slope.

In this case, we require an expression for minimum curve length given a

required SSD.

Crest and sag vertical curves are considered separately.

Stopping Sight Distance and Crest Vertical
Curve Design

Where,

SSD or S = stopping sight distance in m, PVC = point of vertical curve

ℎ1 = height of driver’s eye above PVI = point of vertical intersection

roadway surface in m
PVT = point of vertical tangent
ℎ2 = height of object above
roadway surface in m

L = Length of the curve in m

Stopping Sight Distance and Crest Vertical
Curve Design
To determine the minimum length of curve for a required sight distance,
the properties of a parabola for an equal tangent curve can be used to
show that

For S < L
𝐴𝑆 2
𝐿𝑚 = (12)
200( ℎ1 + ℎ2 )2

For S > L
200( ℎ1 + ℎ2 )2
𝐿𝑚 = 2𝑆 − 𝐴
(13)

Where,

𝐿𝑚 = minimum length of vertical curve in m

𝐴 = absolute value of the difference in grades ( lG1 – G2l )
expressed as a percentage
Stopping Sight Distance and Crest Vertical
Curve Design
Current AASHTO design guidelines (2011) use a driver eye height, ℎ1 , of
1080mm and a roadway object height, ℎ2 , of 600mm.

Substituting AASHTO guidelines for ℎ1 and ℎ2 and letting 𝑆 = SSD, into

eq. 12 and eq. 13 gives

For S < L
𝐴 ∗ 𝑆𝑆𝐷2
𝐿𝑚 = (14)
658

For S > L
658
𝐿𝑚 = 2 ∗ 𝑆𝑆𝐷 − (15)
𝐴

Where,

SSD = stopping sight distance in m

Design Speed and Crest Vertical Curve Design
Eq. 14 and Eq. 15 can be simplified if the initial assumption L > SSD is made,
whereas eq. 14 is always used, giving the linear relationship between A and 𝐿𝑚 .
In this case, we can use:

𝐿𝑚 = KA (16)
Where,

K = horizontal distance, in m, required to effect a 1% change in the slope.

And K is defined as:

𝑆𝑆𝐷2
K= (17)
658

The assumption that L > SSD is used because in many cases L is greater than
SSD and when it is not, using the L > SSD formula gives a longer curve length
and thus the error is minimized.

Note that very short vertical curves are difficult to construct and may not be
warranted for safety purposes. As a result it is common to have minimum curve
length limits that range from 30-100m. Another alternative is to set the minimum
curve length at 3 x design speed (in m and km/hr)
Design Speed and Crest Vertical Curve Design

In practice, policies vary as to how the grade is handled.

The assumption that G = 0 is not really correct.

If we use eq. 12 to calculate for an initial grade of +1.0%, we will

underestimate the stopping sight distance because the vertical curve has
a slope as steeply positive as this only at the PVC.

On the other hand, if we use the final grade to be -2.0%, we will

overestimate the stopping sight distance because the vertical curve has a
slope as steeply negative as this only at the PVT.

As a consequence, some design agencies ignore the effect of grade

completely, while others assume G = 0 for grades <3% and use simple
adjustments for SSD for grades 3% or greater.

In this course, G = 0 is used to calculate the respective equations.

EX3: Design Speed and Crest Vertical Curve
Design
A highway is being designed to AASHTO guidelines with a 120 km/hr
design speed and at one section, an equal tangent vertical curve must be
designed to connect grades of +1.0% and -2.0%.
Determine the minimum length of curve necessary to meet SSD
requirements.

Assume S < L (typically)

𝐴 ∗ 𝑆𝑆𝐷2 3 ∗ 2502
𝐿𝑚 = 658
= 658
= 284.95m

SSD for a speed of 120km/hr from the table is 250m

284.95m > 250m

Therefore, that assumption that L > SSD was correct.

EX4: Stopping Sight Distance and Crest
Vertical Curve Design
If the grades in EX3 intersect at station 3+000, determine the stationing of
the PVC, PVT and curve high point for the minimum curve length based
on SSD requirements.

Using 𝐿𝑚 value from EX3. Since the curve is equal tangent, one half of
the curve will occur before the PVI and the other, after.

PVC is at 3+000 – L/2 = 3+000 – 0+142.5

= 2+857.5

PVT is at 3+000 + L/2 = 3+000 + 0+142.5

= 3+142.5

Use eq. 11 for the stationing of the high point

𝑥ℎ𝑙 = K x lG1l = 95 x 1 = 95m

Or
2+857.5 + 0+095 = 2+952.5
The End