Topic 10 Vertical Curves

Gradients

In the same way that horizontal curves are used to connect intersecting straights in the

horizontal plane, vertical curves are used to connect intersecting straights in the vertical

plane. These straights are usually called gradients and the combination of the of the

gradients and vertical curve is known as vertical alignment.

Gradients are usually expressed as percentages, e.g. 1 in 50 = 2%, 1 in 25% = 4% . As

with transition curves these measurements are directly linked to road design.

In the design calculations the algebraic difference A between the gradients is used. This

requires the introduction of the sign convention that gradients rising in the direction of

increasing chainage are considered to be positive and those falling are considered to he

negative.

This results in six different combinations of gradient. These are shown below.

A = m%(entry gradient) – n% (exit gradient)

The sign convention applied is that gradients which rise in the direction of increasing

chainage are considered positive and gradients which fall in the direction of increasing

chainage are considered negative.

A crest curve is defined as one in which the algebraic difference between the

intersecting gradients is positive.

A sag is defined as one in which the algebraic difference between the intersecting

gradients is negative.

Purpose of Vertical Curves

Vertical curves are similar to horizontal curves in that they are designed for a particular

speed. The main functions of a vertical curve are to provide:

Safety and comfort travelling between gradients

Adequate visibility to enable vehicles to stop or overtake safely

Comfort and safety

For vertical curves, vehicles travelling along at the design speed will experience a radial

force which acts to push the vehicle up away from the road surface for a crest curve or

down into the road surface for a sag curve.

Either can be uncomfortable and dangerous for passengers and these forces are

minimised by restricting the gradients on roadways and by choosing an appropriate type

and length of curve for the situation.

Adequate Visibility

In order that vehicles travelling at the design speed can stop or overtake safely they

must be able to see oncoming vehicles or obstruction clearly and in good time.

The length of vertical curve used must ensure that any visibility requirements are met.

This is achieved using sight distances and K values.

Vertical Curves Geometry

Due to restrictions on the maximum gradients, in practice Vertical curves are

categorised as flat curves.

The definition of a flat curve is that if its length is Lv and radius is R, then:

A vertical curve can be a circular arc, an ellipse or a parabolic curve. In practice a

parabolic curve is often used to achieve a uniform rate of change of gradient and to

gradually introduce the vertical radial force. The equation of the parabola is:

10/1≤R

Lv

2cyx =

Assumptions in Vertical Curve Calculations

Choosing a parabola simplifies the calculations of a vertical curve, further

simplifications are also made if certain assumption are used

Vertical curves are generally designed with equal tangent lengths such that PQ = OR.

Assumptions made are:

Chord PWR = Arc PSR = PQ + QR

Length along the tangents = horizontal length, i.e. PQ = PQ’

QU = QW, i.e. there is no difference in dimensions measured either in the vertical plane

or perpendicular to the entry tangent length

Let QS = e and the length of the curve = Lv. The basic equation is modified to give a

general equation including the design terms from:

Level of Q above P = , i.e. slope x length = the rise

Level of R below Q =

2002100

vv mLLm=

2002100

vv nLLn=

Hence, Level of R above P =

From the assumptions, PW = WR, therefore:

Level of W above P =

From the properties of a parabola

Since QW = Level of Q above P – Level of W above P

In this case the algebraic gradient A = (+m) – (-n) = (m+n)

−=

−

200

)(

200200

vvv LnmnLmL

−

400

)( vLnm

SWQW

QS ==2

800

)(

400

)(

2002

1 vvv LnmLnmmLQS

+=

−−=

800

vALeQS ==

The basic equation of the parabola was x = cy2, So at point Q when y=Lv/2 and x = e:

this gives =>

Therefore

Since then:

This is the general equation of the parabolic vertical curve and it is used to calculate

reduced levels along the curve.

2

2

= vL

ce ( )22/vL

ec =

( )2

2

2/vL

eyx =

800

vALe =

( )vL

Ayx

200

2

=

Reduced Levels on Vertical Curves

The reduced level of any point Z along the vertical curve above is given by the reduced

level of P plus the value of ΔH.

−=∆

vL

AymyH

200100

2

Crest Point and Sag Point

It is often important to know the highest or lowest point on a vertical curve for locating

drainage channels etc. The highest point occurs when ΔH is a maximum. For max or min

value of ΔH, d(ΔH)/dy = 0:

This gives:

And for a maximum of minimum value of ΔH

0100100

)(=

−=

∆

vL

Aym

dy

Hd

VL

Aym

100100=

A

mLy V=

This gives the point along the curve at which a maximum or minimum value occurs. To

find the reduced level it is necessary to substitute back into the reduced levels formula

as follows:

Sight Distances

The length of vertical curve to be used depends on the sight distance. This is the

distance of visibility from one side of the carriageway on which a vertical curve is

located to the other. There are to categories of sight distance:

Stopping Sight Distance (SSD), which is the theoretical forward sight distance required

by a driver to stop safely and comfortably when faced an unexpected hazard.

=

−

=∆

A

mL

A

mL

L

A

A

mLmH

v

V

V

V

200

200100

2

2

22

minmax/

- Full Overtaking Sight Distance (FOSD), which is the length of visibility required by

drivers of vehicles to enable them to overtake other vehicles on the road ahead in safety

and comfort. (FOSD values are greater than SSD values)

It is usually only necessary to design for FOSD at the crest of vertical curves on single

carriageways, as overtaking is not an issue on dual carriageways and on sag curve there

is usually adequate visibility.

For SSD drivers eye level height is taken as 1.05m, tall vehicles are taken as 2.00m and

small objects are taken at 0.26m (i.e. the height of tail lights of a car) which gives the

envelope of visibility below

Typical Sight Distances and K Values

K values

The minimum length of a vertical curve (min LV) for any given road is obtained from the

equation:

Min LV = KA metres

K is a constant obtained from the previous table which are derived from sight distances

and the design speed of the road.

A is the algebraic difference of gradients (the absolute value is used).

K values ensure that the minimum length of a vertical curve, obtained from the

equation above, contains adequate visibility and provides sufficient comfort.

Vertical Curves Example 1: Designing a Vertical Curve

The reduced level of an intersection of a rising gradient of 1.5% and a falling

gradient of 1.0% on a proposed motorway is 93.60m. Given the K value for this

particular road is 55, the through chainage of the intersection is 671.34m and the

vertical curve is to have equal tangent lengths, calculate:

The through chainages of the tangent points of the vertical curve if the minimum

required length is to be used

The reduced level of the tangent points and the reduced levels at exact 20m

multiples of through chainage

The position and reduced level of the highest point on the curve.

Solution:

Through chainage of tangent points:

A = (entry gradient %) – (exist gradient % = (+1.5) – (-1.0) = +2.5%

Min LV = KA = 55 x 2.5 = 137.5 m

Since the minimum length of the curve is to be used and using the assumptions

that the length of the curve along the gradients is the same as along the

horizontal

Through chainage of P

Through chainage of R

Reduced levels of Tangent points and reduced levels at exact 20m multiples of

through chainage along the curve:

Reduced level of P

Reduced level of R

To keep to exact multiples of through chainage there will be the need for an initial

short value of y of 620 - 602.59 = 17.41m. All other values of Y will increase in

steps of 20m until as last short section.

m59.6022

5.13734.671 =

−=

m09.7402

5.13734.671 =

+=

mmLV 57.9203.160.93200

60.93 =−=

−=

mnLV 91.9269.060.93200

60.93 =−=

−=

Working from P towards R the RLs of the points on the curve are as follows:

−+=∆+=

VL

yyHRL

200

5.2

100

5.157.9257.92

2

Reduced level and position of highest point

The highest point along the curve occurs when:

The highest RL is then obtained from:

Highest RL

mA

mLy V 50.82

5.2

5.15.137=

×==

mA

mLRL V

P 19.935.2200

5.15.13757.92

200

22

=

×

×+=

+=

Vertical Curves Example 2: Designing a Vertical curve to pass through a

particular Point.

A parabolic vertical curve having equal tangent lengths is to connect a falling

gradient of 2.40% to a rising gradient of 2.10% on a road designed for a speed of

100 kph. The length of the curve used must at least be enough to ensure that the

minimum K value is 26.

In order to ensure there will be sufficient clearance, the curve must pass through

point Z of through chainage 2781.92m and RL 113.56m. The RL and through

chainage of the intersection point Q are 110.98m and 2815.63m.

Calculate the length of the curve that will meet these requirements.

Solution

the point Z is shown above located a horizontal distance y from the entry point P

and a height x above the entry gradient.

A = (-2.40) – (+2.10) = - 4.50

Since the point Z has a larger through chainage than Q, it must lie beyond the

intersection point. This distance y is given by:

Since x is the vertical offset from the entry gradient to Z:

58.1122

29.562

)63.281592.2871(2

−=

+=−+=

yL

therefore

mLL

y

V

VV

m

LymRLRLx V

QZ

931.3100

)29.5640.2(98.11056.113

100

)2/(

=

×+−=

−+−=

But

= 3.931m

Substituting for LV:

Solving this quadratic gives

( ) vv L

y

L

Ayx

200

50.4

200

22

==

098.1966842.349

)58.1122(200

50.4931.3

2

2

=+=

−=

yy

y

y

my

or

my

y

524.70

895.278

)1(2

)98.19668)(1(442.349[42.3492

=

=

−±=

Substituting into the LV equation:

LV = 455.21m or 28.47m

The minimum LV required is = KA = 26 x 4.50 = 117.00m

Therefore LV = 445.21m