Vertical curve +4% and -3%, through given point

A +4% grade meeting a −3% grade gives a summit curve, $N=0.04+0.03=0.07$.

Stopping sight distance

$V=80$ km/h, $t=2.5$ s, effective friction $=0.35\times0.90=0.315$ (brake efficiency 90%):

$$SSD=0.278(80)(2.5)+\frac{80^2}{254(0.315)}=135.6\ \text{m}$$

Length of curve (assuming $L>S$)

$$L=\frac{NS^2}{4.4}=\frac{0.07(135.6)^2}{4.4}=292.5\ \text{m}$$

Adopt L = 295 m ($>S$, valid); half length = 147.5 m.

RL of PVI from the given point

"60 m right from PVI" is taken as 60 m on the EVC side. Distance of that point from EVC $=147.5-60=87.5$ m. The curve lies below the tangent by $\dfrac{N(x')^2}{2L}$:

$$\text{offset}=\frac{0.07(87.5)^2}{2(295)}=0.908\ \text{m}$$

Tangent level at 60 m right of PVI $=RL_{PVI}-0.03(60)$, and the curve level is this minus the offset:

$$120.10=RL_{PVI}-1.80-0.908\Rightarrow RL_{PVI}=122.808\ \text{m}$$

Levels

$$RL_{BVC}=122.808-0.04(147.5)=116.908\ \text{m},\qquad RL_{EVC}=122.808-0.03(147.5)=118.383\ \text{m}$$

Highest point

$$x=\frac{g_1L}{N}=\frac{0.04(295)}{0.07}=168.57\ \text{m from BVC}$$ $$RL_{max}=RL_{BVC}+\frac{g_1^2L}{2N}=116.908+\frac{0.04^2(295)}{2(0.07)}=120.280\ \text{m}$$

Answer: L = 295 m; RL of BVC = 116.91 m; RL of highest point = 120.28 m; RL of EVC = 118.38 m.