Vertical curve 35 m headlight and momentum grade

Given: valley curve, $L = 35$ m, $g_1 = +3\%$, $g_2 = +9.5\%$, headlight height $h = 0.75$ m, beam angle $\alpha = 1^\circ$, $f = 0.35$, coefficient of inertia $\gamma = 0.85$, exit speed 15 km/h.

$$N = 0.095 - 0.03 = 0.065$$

(i) Design speed

The curve is designed for headlight sight distance (gradient effect neglected), reaction time $t = 2.5$ s. First find $S$.

Try $S<L$: $L = \dfrac{NS^2}{2(h+S\tan\alpha)}$ gives $S = 39.3$ m $> L$, so not valid. Hence $S > L$:

$$L = 2S - \frac{2\,(h + S\tan\alpha)}{N}$$ $$35 = 2S - \frac{2\,(0.75 + 0.017455\,S)}{0.065} = 1.4629\,S - 23.077 \ \Rightarrow\ S = 39.70\ \text{m}$$

$S = 39.7 > L = 35$ m, consistent.

This is the stopping sight distance: $SSD = 0.278\,V\,t + \dfrac{V^2}{254 f}$

$$0.695\,V + \frac{V^2}{88.9} = 39.70\ \Rightarrow\ V = 36.07\ \text{km/h}$$

Design speed $\approx$ 36 km/h (check: SSD at 36 km/h $= 25.0 + 14.6 = 39.6$ m).

(ii) Momentum grade

The vehicle enters the steeper grade (9.5%) at $v_1 = 36/3.6 = 10$ m/s and must leave at $v_2 = 15/3.6 = 4.17$ m/s. Using

$$L = \frac{\gamma\,(v_1^2 - v_2^2)}{2g\,(G+f)}$$

with $G = 0.095$, $f = 0.35$ (taken as the resistance coefficient given):

$$L = \frac{0.85\,(10^2 - 4.17^2)}{2(9.81)(0.095+0.35)} = \frac{0.85\times82.6}{8.731} = 8.05\ \text{m}$$

Answer: design speed $\approx$ 36 km/h; maximum length of the steeper (9.5%) gradient as a momentum grade $\approx$ 8 m. (The value is small because the given friction coefficient 0.35 is large.)