Valley curve design, 100 kmph, 1 in 40

Given: $V = 100$ km/h, $g_1 = -1/40 = -2.5\%$, $g_2 = +1/30 = +3.33\%$, BVC elevation $= 312.56$ m.

Assumptions (IRC/NRS): reaction time 2.5 s, $f = 0.35$, comfort rate $C = 0.6$ m/s³, headlight height 0.75 m, beam angle $1^\circ$.

$$N = 0.025 + 0.03333 = 0.05833$$

$v = 100/3.6 = 27.78$ m/s.

1. Comfort condition

$$L = 2\sqrt{\frac{N v^3}{C}} = 2\sqrt{\frac{0.05833\times 27.78^3}{0.6}} = 2\sqrt{2083.8} = 91.3\ \text{m}$$

2. Headlight sight distance

SSD:

$$SSD = 0.278(100)(2.5) + \frac{100^2}{254\times0.35} = 69.5 + 112.5 = 182.0\ \text{m}$$

Assuming $S < L$:

$$L = \frac{N S^2}{2\,(h + S\tan\alpha)} = \frac{0.05833\,(182.0)^2}{2\,(0.75 + 182.0\tan1^\circ)} = \frac{1932.2}{2(3.927)} = 246.0\ \text{m}$$

$S = 182 < 246$ so the assumption holds.

3. Design length

Adopt the larger: $L = 246$ m, rounded to $L = 250$ m.

4. Lowest point

Distance from BVC:

$$x = \frac{g_1 L}{N} = \frac{0.025\times250}{0.05833} = 107.1\ \text{m}$$

Elevation (curve equation $y = BVC + g_1x + \dfrac{N x^2}{2L}$ with $g_1 = -0.025$):

$$\begin{aligned} RL_{low} &= 312.56 - 0.025(107.14) + \frac{0.05833(107.14)^2}{2(250)} \\ &= 312.56 - 2.679 + 1.339 = 311.22\ \text{m} \end{aligned}$$

5. End of curve (EVC)

$$RL_{EVC} = 312.56 + \frac{(g_1+g_2)L}{2} = 312.56 + \frac{(-0.025+0.03333)(250)}{2} = 313.60\ \text{m}$$

The EVC is at 250 m from the BVC.

Answer: $L = 250$ m (comfort 91.3 m, headlight 246.0 m); lowest point is 107.1 m from BVC at RL 311.22 m; end of curve at 250 m from BVC with RL 313.60 m.