Module 3: Geometric design of highways

Lecture 17 Vertical alignment I

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Case b. Length of summit curve less than sight distance

The second case is illustrated in figure 1

Figure 1: Length of summit curve (L<S)

From the basic geometry, one can write

$$S=\frac{L}{2}+\frac{h_1}{n_1}+\frac{h_2}{n_2} =\frac{L}{2}+\frac{h_1}{n_1}+\frac{h_2}{N-n_2}$$ (1)

Therefore for a given $L$, $h_1$ and $h_2$ to get minimum $S$, differentiate the above equation with respect to $h_1$ and equate it to zero. Therefore,

$$\frac{dS}{dh_1} = \frac{-h_1}{n_1^2} + \frac{h_2}{{N-{n_1}}^2} = 0$$ (2)

Solving for $n_1$,

$$n_1 = \frac{N\sqrt{h_1 h_2} - h_1N}{h_2 - h_1}$$ (3)

Now we can substitute $n$ back to get the value of minimum value of $L$ for a given $n_1$, $n_2$, $h_1$ and $h_2$. Therefore,

$$S = \frac{L}{2}+\frac{{h_1}}{\frac{N\sqrt{h_1 h_2} - h_1N}{h... ...1}}+\frac{{h_2}}{{N-\frac{N\sqrt{h_1 h_2} - h_1N}{h_2 - h_1}}}$$ (4)

Solving for $L$,

$$L=2S-\frac{\left(\sqrt{2h_1}+\sqrt{2h_2}\right)^2}{N}$$ (5)

When stopping sight distance is considered the height of driver's eye above the road surface ($h_1$) is taken as 1.2 metres, and height of object above the pavement surface ($h_2$) is taken as 0.15 metres. If overtaking sight distance is considered, then the value of driver's eye height ($h_1$) and the height of the obstruction ($h_2$) are taken equal as 1.2 metres.