Highway Engineering

Geometric Alignment and Design 177

Example 6.5 Contd

Shift:
Using Equation 6.24:

S=L^2 /24R=(86.03)^2 ∏ 24 ¥ 510

=0.605 m

Length of IT:
Using Equation 6.29

Form of the transition curve:
Using Equation 6.31
x =y^3 ∏ 6 RL
=y^3 ∏ 6 ¥ 510 ¥86.03
=y^3 ∏263 251.8 (6.32)

Co-ordinates of point at which circular arc commences:
This occurs where y equals the transition length (86.03 m).
At this point, using Equation 6.31:

x =(86.03)^2 ∏(6 ¥510)

=2.419 m

This point can now be fixed at both ends of the circular arc. Knowing its
radius we are now in a position to plot the circle.

Note: In order to actually plot the curve, a series of offsets must be gener-
ated. The offset length used for the intermediate values of y is typically
between 10 and 20 m. Assuming an offset length of 10 m, the values of x at
any distance y along the straight joining the tangent point to the intersection
point, with the tangent point as the origin (0,0), are as shown in Table 6.10,
using Equation 6.32.

IT

m

=+()()+

=()()+

=+

=

RStan / L/

. tan /. /
..
.

q 22

510 605 42 2 86 03 2

196 00 43 015

239 015

yx

10 0.0038

20 0.0304

30 0.1026

40 0.2431

50 0.4748

60 0.8205

70 1.303

80 1.945

Table 6.10Offsets at

10 m intervals