Vertical curve, -3% and 1.5% at chainage 2000

A -3% grade meeting a +1.5% grade forms a valley (sag) curve. Its length is the larger of the headlight-sight-distance length and the comfort length (IRC / NRS 2070, parabolic curve).

Data

$g_1=-3\%,\ g_2=+1.5\%$, so $N=0.03+0.015=0.045$. $V=100$ km/h, $f=0.35$, reaction time $t=2.5$ s, headlight height $h_1=0.75$ m, beam angle $\alpha=1^\circ$. PVI at chainage 2000 m, RL 500.00 m.

Stopping sight distance

$$SSD=0.278Vt+\frac{V^2}{254f}=0.278(100)(2.5)+\frac{100^2}{254(0.35)}=182.0\ \text{m}$$

Length of curve

Headlight criterion (assuming $L>S$), with $2\tan1^\circ = 0.0349$:

$$L=\frac{NS^2}{2(h_1+S\tan\alpha)}=\frac{0.045(182.0)^2}{1.5+0.0349(182.0)}=189.8\ \text{m}$$

$L>S$, so the assumption holds. Comfort criterion (rate of change of centrifugal acceleration $C=0.6$ m/s³, $v=27.78$ m/s):

$$L=2\sqrt{\frac{Nv^3}{C}}=80.2\ \text{m}$$

Adopt L = 190 m (the headlight criterion governs).

Tangent points

Half length = 95 m.

• BVC chainage = 2000 − 95 = 1905 m; RL = 500 + 0.03(95) = 502.85 m
• EVC chainage = 2000 + 95 = 2095 m; RL = 500 + 0.015(95) = 501.425 m

Lowest point

Distance from BVC where the slope is zero:

$$x=\frac{|g_1|L}{N}=\frac{0.03(190)}{0.045}=126.67\ \text{m}$$

Chainage = 1905 + 126.67 = 2031.67 m. RL of the lowest point (with $g_1 = -0.03$):

$$y=RL_{BVC}+g_1x+\frac{Nx^2}{2L}=502.85-0.03(126.67)+\frac{0.045(126.67)^2}{2(190)}=500.950\ \text{m}$$

Answer: L = 190 m; BVC at 1905 m (RL 502.85 m); EVC at 2095 m (RL 501.425 m); lowest point at chainage 2031.67 m, RL 500.950 m.