Highway Engineering

188 Highway Engineering

Example 6.8 Contd

Thus the required crest curve length more than doubles in value if the object

height is reduced to zero.

L==S ¥

¥

=

A

2H

0.065 160

2 1.05

792 m

2

1

2

Example 6.9

Using the same basic data as Example 6.8, but with the following straight-

line gradients:

p =+0.02

q =-0.02

calculate the required curve length assuming a motorist’s eye height of

1.05 m and an object height of 0.26 m.

Solution

In this case:

Given that in this case e <H 1 ,S>Las the sight distance is greater than the

curve length.

Therefore, using Equation 6.49:

Note: if the object height is reduced to zero, then the required curve length

is calculated from Equation 6.64:

LS=- =¥ -

¥

=-

=

2

2H

A

2 160

2 1.05

0.04

320 52.5

267.5 m

1

LSm^12

22

2

2H H

A

2 160

2 1.05 0.26

0.04

320 117.75

202.25 m

=-

[]+

=¥ -

[]+

=-

=

eqp

8

0.02 0.02 160 8

0.8 m

=- -()=--()- ¥∏

=

L