Vertical curve for overtaking sight distance

Given: $g_1 = +1.2\%$, $g_2 = -1.08\%$, two-lane two-way road, $V = 60$ km/h, acceleration $A = 2.5$ km/h/s, BVC elevation $= 334.792$ m.

$$N = 0.012 + 0.0108 = 0.0228$$

Assumptions (IRC): reaction time $t = 2$ s; speed of the overtaken vehicle $v_b = V - 16 = 44$ km/h $= 12.22$ m/s; spacing $s = 0.7v_b + 6$ with $v_b$ in m/s (equivalently $0.2V_b + 6$ with $V_b$ in km/h); eye height and object height both 1.2 m for overtaking.

$v = 60/3.6 = 16.67$ m/s.

Step 1: Overtaking sight distance

$$\begin{aligned} d_1 &= v_b\,t = 12.22\times2 = 24.4\ \text{m} \\ s &= 0.7(12.22) + 6 = 14.56\ \text{m} \\ T &= \sqrt{\frac{14.4\,s}{A}} = \sqrt{\frac{14.4\times14.56}{2.5}} = 9.16\ \text{s} \\ d_2 &= v_bT + 2s = 12.22(9.16) + 2(14.56) = 141.0\ \text{m} \\ d_3 &= vT = 16.67\times9.16 = 152.6\ \text{m} \end{aligned}$$

(The factor 14.4 is $4\times3.6$, because $A$ is in km/h/s and $s$ in metres.)

$$OSD = 24.4 + 141.0 + 152.6 = 318.1\ \text{m}$$

Step 2: Length of summit curve

Try $S < L$: $L = \dfrac{NS^2}{9.6} = \dfrac{0.0228(318.1)^2}{9.6} = 240.3$ m, which is less than $S$, so the assumption fails. Use $S > L$:

$$L = 2S - \frac{9.6}{N} = 2(318.1) - \frac{9.6}{0.0228} = 636.1 - 421.1 = 215.1\ \text{m}$$

$S = 318.1 > L = 215.1$ m, consistent. $L = 215.1$ m.

Step 3: Highest point

$$x = \frac{g_1L}{N} = \frac{0.012\times215.1}{0.0228} = 113.2\ \text{m from BVC}$$ $$RL_{high} = 334.792 + 0.012(113.2) - \frac{0.0228\,(113.2)^2}{2\times215.1} = 334.792 + 1.358 - 0.679 = 335.47\ \text{m}$$

Step 4: End of curve (EVC)

$$RL_{EVC} = 334.792 + \frac{(0.012-0.0108)(215.1)}{2} = 334.792 + 0.129 = 334.92\ \text{m}$$

The EVC is 215.1 m from the BVC.

Answer: OSD $= 318.1$ m; $L = 215.1$ m; highest point 113.2 m from BVC at RL $335.47$ m; end of curve at RL $334.92$ m.