Highway Engineering

[Mg¥Sin(a)] +P=[(M¥v^2 /R) ¥Cos (a)] (6.7)

The side frictional force,P, can be expressed as:

(6.8)

(mis defined as the side friction factor)

Substituting Equation 6.8 into Equation 6.7, the following expression is derived:

Dividing across by MgCos(a):

(6.9)

If we ignore the term mv^2 /gRtan(a) on the basis that it is extremely small, the
following final expression is derived:

The term tan(a) is in fact the superelevation e.If in addition we express veloc-
ity in kilometres per hour rather than metres per second, and given that gequals
9.81 m/s^2 , the following generally used equation is obtained:

(6.10)

This expression is termed the minimum radius equation. It is the formula which
forms the basis for the values ofR illustrated in Table 6.9.
In UK design practice, it is assumed that, at the design speed, 55% of the cen-
trifugal force is balanced by friction, with the remaining 45% being counteracted
by the crossfall. Thus, Equation 6.10 becomes:

(6.11)

Therefore:

(6.12)

Therefore, assuming ehas a value of 5% (appropriate for the desirable minimum
radius R):

R=0.07069V^2 (6.13)

Taking a design speed of 120 km/hr:

R=0.07069(120)^2

=1018 m

R

V

e

=

0 353.^2

e

V

R

e

V

R

e

=

¥

= ()

045

127

0 353

2

2

.

.

or

is expressed in percentage terms

V

R

e

2

127

=+m

tan()am+=vgR^2

tan()amm++vgR^22 tan()a=vgR

[]Mg¥Sin()am+¥[]Mg Cos()a+¥ ¥M v R^22 Sin()a=¥[]()M v R¥Cos()a

PW C

Mg M v R

=¥[]()+¥()

=¥[]()+¥ ¥()

ma a

ma a

Cos Sin

Cos^2 Sin

Geometric Alignment and Design 169