Valley curve design, 80 kmph, 1 in 40

Reading: the RL of the beginning of the curve is printed as "2.12.86" in the scan; I have read it as 212.86 m. For any other value, simply add the difference to every level below.

A descending 1 in 40 grade meeting an ascending 1 in 30 grade forms a valley curve.

Data

$g_1=-\dfrac{1}{40}=-2.5\%$, $g_2=+\dfrac{1}{30}=+3.33\%$, $N=\dfrac{1}{40}+\dfrac{1}{30}=0.05833$. $V=80$ km/h, $f=0.35$, $t=2.5$ s, $C=0.6$ m/s³, headlight height 0.75 m, beam angle 1° (IRC/NRS).

Length of curve

$$SSD=0.278(80)(2.5)+\frac{80^2}{254(0.35)}=127.6\ \text{m}$$

• Comfort: $L=2\sqrt{\dfrac{NV^3}{C}}=65.3$ m
• Headlight: $L=\dfrac{NS^2}{1.5+0.035S}=\dfrac{0.05833(127.6)^2}{1.5+0.035(127.6)}=159.2$ m ($L>S$, valid)

Adopt L = 160 m (half = 80 m).

Levels

$$RL_{BVC}=212.86\ \text{m}$$ $$RL_{PVI}=212.86-\frac{1}{40}(80)=210.860\ \text{m}$$ $$RL_{EVC}=210.860+\frac{1}{30}(80)=213.527\ \text{m}$$

Lowest point

$$x=\frac{g_1L}{N}=\frac{160/40}{0.05833}=68.57\ \text{m from BVC}$$ $$RL_{low}=212.86-\frac{68.57}{40}+\frac{0.05833(68.57)^2}{2(160)}=212.003\ \text{m}$$

Answer: L = 160 m; beginning RL = 212.86 m; lowest point RL = 212.00 m at 68.6 m from BVC; end RL = 213.53 m.