Superelevation

• Superelevation is the rotation of the pavement on the approach to and through a horizontal curve. It is intended to assist the driver by counteracting the lateral acceleration produced by tracking the curve.

• ratio of the pavement slope to width, ranging from 0 to 0.12 foot/feet..

• Denoting the centrifugal force by C, we have from the theory of mechanics:

W=the weight of loaded car in poundsV= velocity in feet per secondR= radius of curve in feet.

2W vC

g R

æ ö÷ç ÷= ç ÷ç ÷çè ø

WӨ

therefore,

2

2

W vW e

g R

ve

gR

* = *

=

C Wtan W e= q= *

• In metric units, g= 9.81 m/sec2

2

2

2

1000Kv m/ s

3600(1000K)

e3600 9.81 R

0.0079Ke

R

=

=* *

=

Curve Resistance

-it is when a vehicle takes a curve, external forces act on the front wheels of the vehicle. These forces have components that retard the forward motion of the vehicle. This resistance depends on the radius of curvature and the speed of the vehicle.

Example:

Radius of Curvature• Vertical curves on roads are parabolic,• Horizontal curves are based on circles.

When a vehicles moves around a horizontal curve, it is subject to the outward radial force ( centrifugal force). The inward force is not due to gravity, but rather because of the friction between tires and the roadway.

Computing Radius of Curvature

• Let tan Ө =e, g=9.81 m/sec2

u is in km/hr

A spiral 80 meters long connects a tangent with a 6030’ circular curve. The gauge of the track on the curve is 1.5 m. Find the elevation of the outer rail at the midpoint if the velocity of the fastest bus to pass over the curve is 6o kph.

2

cc

2

2

2

0.0079Ke

R

0.0079(60)(1.5)

176.39e 0.241m

.241e

2

=

=

=

=