Handbook of Civil Engineering Calculations

from tangent through A = 4.66 ft (1.4203 m); 4.66/11.232 = 0.415; (208 + 60)

2

/(416)

2

=

0.415. This is acceptable.

SfGHT DISTANCE ON A VERTICAL CURVE

A vertical summit curve has tangent grades of+2.6 and -1.5 percent. Determine the min-
imum length of curve that is needed to provide a sight distance of 450 ft (137.2 m) to an
object 4 in (101.6 mm) in height. Assume that the eye of the motorist is 4.5 ft (1.37 m)
above the roadway.

Calculation Procedure:

1. State the equation for minimum length when S < L
   The vertical curvature of a road must be limited to ensure adequate visibility across the
   summit. Consequently, the distance across which a given change in grade may be effected
   is subject to a lower limit imposed by the criterion of sight distance.
   Let S denote the required sight distance and L the minimum length of curve. In Fig.
   23, let E denote the position of the motorist's eye and P the top of an object. Assume that
   the curve has the maximum allowable curvature, so that the distance from E to P equals S.
   Applying Eq. 44a gives

A&

J = ~£ (Afa

100[(2/J 1 )

172

+ ^

172

]

2 V ;

where A = Ga - Gb, in percent.

FIGURE 23. Visibility on vertical summit curve.