Vertical curve -2.5% and +3.5%, 1268.50 m

Data: $g_1 = -2.5\% = -0.025$, $g_2 = +3.5\% = +0.035$ (a valley or sag curve), IP at chainage 780.00 m, RL 1267.00 m. The curve passes through RL 1268.50 m at chainage 780 m (above the IP, as it must for a sag curve).

Length of the curve

$e = 1268.50 - 1267.00 = 1.50$ m, and for an equal tangent parabolic curve $e = \dfrac{(g_2 - g_1)L}{8}$:

$$L = \frac{8e}{g_2 - g_1} = \frac{8\times 1.50}{0.035 + 0.025} = 200.00\ \text{m}$$

Elements

• PVC chainage $= 780 - 100 = 680.00$ m; PVT chainage $= 880.00$ m
• RL of PVC $= 1267.00 + 0.025\times 100 = 1269.500$ m; RL of PVT $= 1267.00 + 0.035\times 100 = 1270.500$ m
• $r = \dfrac{g_2 - g_1}{L} = 0.000300$ per m

Levels at full stations (30 m pegs)

$$RL_x = RL_{PVC} + g_1x + \frac{r}{2}x^2$$

Lowest point

$$x = \frac{-g_1}{r} = \frac{0.025}{0.000300} = 83.33\ \text{m}$$

Chainage $= 680.00 + 83.33 = 763.33$ m and RL $= 1268.458$ m.

Check: level at chainage 780 m (x = 100 m) $= 1268.500$ m = 1268.500 m as required.

Answer: L = 200 m; lowest point at chainage 763.33 m, RL 1268.458 m.