Vertical curve -3.25% and +3.75%, 225 m

A −3.25% grade meeting +3.75% is a valley curve; $N=0.0325+0.0375=0.07$.

Length ($S=225$ m)

Headlight height and beam angle are not given, so the standard values $h_1=0.75$ m and $\alpha=1^\circ$ (IRC/NRS) are assumed:

$$L=\frac{NS^2}{1.5+0.035S}=\frac{0.07(225)^2}{1.5+0.035(225)}=378.0\ \text{m}$$

$L>S$, valid. Adopt L = 380 m (half = 190 m).

RL of EVC from the given point

The point is 18 m before EVC. Measured back from EVC along the +3.75% tangent, the curve lies above the tangent by $\dfrac{Nx^2}{2L}$:

$$125.32=RL_{EVC}-0.0375(18)+\frac{0.07(18)^2}{2(380)}\Rightarrow RL_{EVC}=125.32+0.675-0.030=125.965\ \text{m}$$

RL of PVI and BVC

$$RL_{PVI}=RL_{EVC}-0.0375(190)=118.840\ \text{m},\qquad RL_{BVC}=RL_{PVI}+0.0325(190)=125.015\ \text{m}$$

Lowest point

$$x=\frac{g_1L}{N}=\frac{0.0325(380)}{0.07}=176.43\ \text{m from BVC}$$ $$RL_{low}=125.015-0.0325(176.43)+\frac{0.07(176.43)^2}{2(380)}=122.148\ \text{m}$$

Answer: RL of BVC = 125.02 m; lowest point = 122.15 m (at 176.4 m from BVC); RL of EVC = 125.97 m.