Vertical curve with intermediate sight distance

Given: ISD $= 255$ m, $V = 80$ km/h, $f = 0.35$ (the ISD is $2\times SSD$: $2\times127.6 = 255.2$ m, which is consistent with these values), PVI at chainage $1+000$, RL $500$ m, $g_1 = +5\%$.

Reading: a highest point exists only if the second grade falls, so the grades are $g_1 = +5\%$ and $g_2 = -1.5\%$. This is a summit curve with

$$N = 0.05 + 0.015 = 0.065$$

Length of the curve

Intermediate sight distance, eye height $H = 1.2$ m and object height $h = 1.2$ m (IRC): $(\sqrt{2H}+\sqrt{2h})^2 = 9.6$. Assuming $S < L$:

$$L = \frac{N S^2}{9.6} = \frac{0.065\times255^2}{9.6} = 440.3\ \text{m}$$

$S = 255 < 440.3$ so the assumption holds.

(a) Beginning of the curve (BVC)

$$\begin{aligned} \text{Chainage} &= 1000 - \frac{L}{2} = 1000 - 220.14 = 779.86\ \text{m}\ (0+779.86)\\ RL &= 500 - 0.05\times220.14 = 488.99\ \text{m} \end{aligned}$$

(b) Highest point

Distance from BVC:

$$x = \frac{g_1 L}{N} = \frac{0.05\times440.27}{0.065} = 338.67\ \text{m}$$

Chainage $= 779.86 + 338.67 = 1118.53$ m.

$$RL = 488.99 + 0.05(338.67) - \frac{0.065\,(338.67)^2}{2\times440.27} = 488.99 + 16.93 - 8.47 = 497.46\ \text{m}$$

(c) Point 50 m to the right of PVI

Chainage $= 1050$ m, and $x = 220.14 + 50 = 270.14$ m from BVC:

$$RL = 488.99 + 0.05(270.14) - \frac{0.065\,(270.14)^2}{2\times440.27} = 488.99 + 13.51 - 5.39 = 497.11\ \text{m}$$

Answer: BVC at chainage 779.86 m (RL 488.99 m); highest point at chainage 1118.53 m (RL 497.46 m); point 50 m right of PVI (chainage 1050 m) RL 497.11 m. End of curve at chainage 1220.14 m, RL 496.70 m.