Vertical curve +3.5% and -2.75%, 30 m pegs

Data: $g_1 = +3.5\% = +0.035$, $g_2 = -2.75\% = -0.0275$ (summit curve), IP at chainage 2800.00 m, RL 1470.00 m. The curve must pass through RL 1468.75 m at chainage 2800.00 m, i.e. at the IP vertical.

Length of the curve

The offset from the IP to the curve at mid-point is $e = 1470.00 - 1468.75 = 1.25$ m. For an equal-tangent parabolic curve

$$e = \frac{(g_1 - g_2)L}{8} \quad\Rightarrow\quad L = \frac{8e}{g_1 - g_2} = \frac{8\times 1.25}{0.035 + 0.0275} = 160.00\ \text{m}$$

Each tangent length $= L/2 = 80.00$ m.

Elements

• PVC chainage $= 2800 - 80 = 2720.00$ m; PVT chainage $= 2880.00$ m
• RL of PVC $= 1470.00 - 0.035\times 80 = 1467.200$ m; RL of PVT $= 1470.00 - 0.0275\times 80 = 1467.800$ m
• $r = \dfrac{g_2 - g_1}{L} = -0.0003906$ per m

Levels at the full stations (30 m pegs)

$$RL_x = RL_{PVC} + g_1 x + \frac{r}{2}x^2$$

Highest point

$$x = \frac{-g_1}{r} = \frac{0.035}{0.0003906} = 89.60\ \text{m}$$

Chainage $= 2720.00 + 89.60 = 2809.60$ m, and RL $= 1467.200 + 0.035(89.60) + \frac{-0.0003906}{2}(89.60)^2 = 1468.768$ m.

Check: level at the IP chainage (x = 80 m) $= 1468.750$ m = 1468.750 m as required.

Answer: L = 160 m; highest point at chainage 2809.60 m, RL 1468.768 m.