Vertical curve 5% and 4% at 50+050

An ascending grade of +5% followed by a descending grade of −4% gives a summit curve.

Data

$g_1=+5\%,\ g_2=-4\%$, $N=0.05+0.04=0.09$, $R=10000$ m, PVI at chainage 50+050 ($=50\,050$ m), RL 500.50 m.

Length and tangent points

$$L=NR=0.09(10000)=900\ \text{m},\qquad \frac{L}{2}=450\ \text{m}$$

• BVC chainage = 50 050 − 450 = 49600 m (49+600)
• EVC chainage = 50 050 + 450 = 50500 m (50+500)

Elevations of BVC and EVC

$$RL_{BVC}=500.50-0.05(450)=478.00\ \text{m},\qquad RL_{EVC}=500.50-0.04(450)=482.50\ \text{m}$$

Elevation on the curve

With $x$ measured from BVC: $y=RL_{BVC}+g_1x-\dfrac{x^2}{2R}$.

• 300 m before PVI: $x=450-300=150$ m (chainage 49+750). $y=478.00+0.05(150)-\dfrac{150^2}{20000}=\mathbf{484.375}$ m
• At the PVI: $x=450$ m (chainage 50+050). $y=478.00+0.05(450)-\dfrac{450^2}{20000}=\mathbf{490.375}$ m
• 300 m after PVI: $x=450+300=750$ m (chainage 50+350). $y=478.00+0.05(750)-\dfrac{750^2}{20000}=\mathbf{487.375}$ m

Check at EVC ($x=900$): $y=482.50$ m, which equals the RL of EVC. The points 600 m from the PVI lie beyond the curve, on the tangents: RL $=500.50-0.05(600)=470.50$ m at 49+450 and $500.50-0.04(600)=476.50$ m at 50+650.

Answer: BVC at 49+600 (RL 478.00 m), EVC at 50+500 (RL 482.50 m); elevation 300 m before PVI = 484.375 m, on the curve at the PVI = 490.375 m, and 300 m after PVI = 487.375 m.