Valley curve 1/35 and 1/45, 80 kmph

A descending 1 in 35 grade meeting an ascending 1 in 45 grade gives a valley curve.

Data

$g_1=-\dfrac{1}{35}=-2.857\%$, $g_2=+\dfrac{1}{45}=2.222\%$, so $N=\dfrac{1}{35}+\dfrac{1}{45}=0.05079$. $V=80$ km/h, $f=0.35$, $t=2.5$ s, $C=0.6$ m/s³, headlight height 0.75 m, beam angle 1° (IRC/NRS).

Stopping sight distance

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

Length of valley curve

• Comfort: $L=2\sqrt{\dfrac{NV^3}{C}}=2\sqrt{\dfrac{0.05079(22.22)^3}{0.6}}=61.0$ m
• Headlight: $L=\dfrac{NS^2}{1.5+0.035S}=\dfrac{0.05079(127.6)^2}{1.5+0.035(127.6)}=138.6$ m ($L>S$, valid)

Adopt L = 140 m (headlight criterion governs and also satisfies comfort).

Levels (PVI RL = 212.36 m, half length = 70 m)

$$RL_{BVC}=212.36+\frac{1}{35}(70)=214.360\ \text{m}$$ $$RL_{EVC}=212.36+\frac{1}{45}(70)=213.916\ \text{m}$$

Lowest point at $x=\dfrac{g_1L}{N}=\dfrac{(140)/35}{0.05079}=78.75$ m from BVC:

$$RL_{low}=214.360-\frac{78.75}{35}+\frac{0.05079(78.75)^2}{2(140)}=213.235\ \text{m}$$

Apex distance and mid-ordinate

The vertical distance from the PVI (apex) to the curve, which also equals the mid-ordinate of the curve above/below its chord, is

$$E=\frac{NL}{8}=\frac{0.05079(140)}{8}=0.889\ \text{m}$$

(The curve lies 0.889 m above the PVI, at mid-length, level 213.249 m.)

Answer: L = 140 m; RL of BVC = 214.36 m; lowest point = 213.24 m; RL of EVC = 213.92 m; apex distance = mid-ordinate = 0.89 m.