Module 3: Geometric design of highways

Lecture 16 Horizontal alignment III

1

2

3

4

5

6

7

8

9

10

1. Rate of change of centrifugal acceleration

At the tangent point, radius is infinity and hence centrifugal acceleration is zero. At the end of the transition, the radius R has minimum value R. The rate of change of centrifugal acceleration should be adopted such that the design should not cause discomfort to the drivers. If $c$ is the rate of change of centrifugal acceleration, it can be written as:

$$c = \frac{\frac{v^2}{R}-0}{t},$$

$$= \frac{\frac{v^2}{R}}{\frac{L_s}{v}},$$

$$= \frac{v^3}{L_sR}.$$

Therefore, the length of the transition curve $L_{s_1}$ in $m$ is

$$L_{s_1}=\frac{v^3}{cR},$$ (1)

where $c$ is the rate of change of centrifugal acceleration given by an empirical formula suggested by by IRC as below:

$$c = \frac{80}{75+3.6v},$$ (2)

$${\mathrm subject~to:~~~~~~~~~~~~~~~~~~}$$

$$c_{{\mathrm min}} = 0.5,$$

$$c_{{\mathrm max}} = 0.8.$$

2. Rate of introduction of super-elevation

Raise ($E$) of the outer edge with respect to inner edge is given by $E=eB=e(W+W_e)$. The rate of change of this raise from $0$ to $E$ is achieved gradually with a gradient of $1$ in $N$ over the length of the transition curve (typical range of $N$ is 60-150). Therefore, the length of the transition curve $L_{s_2}$ is:

$$L_{s_2}=Ne(W+W_e)$$ (3)

3. By empirical formula

IRC suggest the length of the transition curve is minimum for a plain and rolling terrain:

$$L_{s_3}=\frac{35v^2}{R}$$ (4)

and for steep and hilly terrain is:

$$L_{s_3}=\frac{12.96v^2}{R}$$ (5)

and the shift $s$ as:

$$s=\frac{L_s^2}{24R}$$ (6)

The length of the transition curve $L_s$ is the maximum of equations 1, 3 and 4or5, i.e.

$$L_s={\mathrm Max:~}(L_{s_1},L_{s_2},L_{s_3})$$ (7)