Vertical curve 200 m, +1.50% then -2.5%

Data: $g_1 = +1.5\% = +0.015$, $g_2 = -2.5\% = -0.025$, $L = 200$ m (equal tangents of 100 m), RL of the point of intersection (IP) = 1185.795 m. The chainage of the IP is not given; it is assumed to be 1000.000 m (0+1000), so that the PVC is at 900 m and the curve pegs at even 20 m stations fall at 900, 920, ... 1100 m. It is a summit (crest) curve.

        g1=+1.5%     IP     g2=-2.5%
              ___.-'`'-.___
         PVC./   (crest)    \.PVT
         900                  1100

Elements

• Rate of change of grade: $r = \dfrac{g_2 - g_1}{L} = \dfrac{-0.025 - 0.015}{200} = -0.000200$ per m
• PVC chainage $= 1000 - 100 = 900.00$ m; PVT chainage $= 1100.00$ m
• RL of PVC $= 1185.795 - 0.015\times 100 = 1184.295$ m
• RL of PVT $= 1185.795 - 0.025\times 100 = 1183.295$ m

Level on the curve (parabola)

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

where $x$ is the distance from the PVC.

Highest point

The curve is highest where the grade is zero: $g_1 + r x = 0$.

$$x = \frac{-g_1}{r} = \frac{0.015}{0.0002} = 75.00\ \text{m}$$

Chainage $= 900.00 + 75.00 = 975.00$ m and

$$RL = 1184.295 + 0.015\times 75.00 - 0.0001\times 75.00^2 = 1184.858\ \text{m}$$

Check: the level at mid-curve (x = 100 m) is 1184.795 m, which is midway between the IP and the mid-point of the chord PVC-PVT (1183.795 m), i.e. $e = \dfrac{(g_1-g_2)L}{8} = 1.00$ m below the IP.

Answer: highest point at chainage 975.00 m with RL 1184.858 m; the levels at the 20 m stations are given in the table.