Vertical curve with 300 m grade sections

An ascending 4% grade followed by a descending 3% grade gives a summit curve, $N=0.04+0.03=0.07$.

Stopping sight distance

$V=100$ km/h, $t=2.5$ s, coefficient of friction $f=0.35$ (assumed, IRC value for 100 km/h), so the effective $f=0.35\times0.90=0.315$:

$$SSD=0.278(100)(2.5)+\frac{100^2}{254(0.315)}=194.5\ \text{m}$$

Length of curve

$$L=\frac{NS^2}{4.4}=\frac{0.07(194.5)^2}{4.4}=601.7\ \text{m}$$

The grade sections are each 300 m long, so the curve can extend at most 300 m on each side of the PVI. Adopt L = 600 m (half = 300 m). The sight distance this gives, $S=\sqrt{4.4L/N}=194.2$ m, is practically the required SSD (194.5 m), so the requirement is met.

Formation level of the PVI (from the given point)

The point 25 m right of PVI is $300-25=275$ m from EVC. Offset below the tangent:

$$y_o=\frac{N(275)^2}{2L}=\frac{0.07(275)^2}{2(600)}=4.411\ \text{m}$$ $$120.105=RL_{PVI}-0.03(25)-4.411\Rightarrow RL_{PVI}=125.266\ \text{m}$$

Main levels

$$RL_{BVC}=125.266-0.04(300)=113.266\ \text{m},\qquad RL_{EVC}=125.266-0.03(300)=116.266\ \text{m}$$

• 50 m before PVI ($x=250$ m from BVC): offset $=\dfrac{0.07(250)^2}{1200}=3.646$ m; level $=125.266-0.04(50)-3.646=\mathbf{119.621}$ m
• 50 m after PVI: level $=125.266-0.03(50)-3.646=\mathbf{120.121}$ m
• Highest point: $x=\dfrac{g_1L}{N}=\dfrac{0.04(600)}{0.07}=342.86$ m from BVC

$$RL_{max}=113.266+\frac{0.04^2(600)}{2(0.07)}=120.124\ \text{m}$$

Answer: RL of PVI = 125.27 m; beginning (BVC) = 113.27 m; end (EVC) = 116.27 m; highest point = 120.12 m; 50 m before PVI = 119.62 m; 50 m after PVI = 120.12 m.