--- title: "Vertical curve 5% and 4% at 50+050" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-78p-vc-5-4" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-78p-vc-5-4 site: BCT Engineering --- # Vertical curve 5% and 4% at 50+050 Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2078 Poush, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-78p-vc-5-4 - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question An ascending gradient meets the descending gradient: the grades are 5% and 4% respectively. Locate the chainage at the beginning and end of vertical curve. Calculate the elevation of the road at every 300 m on either side of the point of vertical intersection (PVI). If the elevation and chainage of the PVI is 500.50 m and 50+050 km respectively. Assume R = 10000 m. ## Short answer BVC at chainage 49+600 (RL 478.00 m), EVC at 50+500 (RL 482.50 m). Elevations: 300 m before PVI 484.375 m, at PVI 490.375 m, 300 m after PVI 487.375 m. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A +5% grade meeting a -4% grade forms which curve and what is N? A. Summit curve, N = 0.09 B. Valley curve, N = 0.09 C. Summit curve, N = 0.01 D. Valley curve, N = 0.01 Answer: A. An upgrade followed by a downgrade is a summit; N is the algebraic difference 0.05 + 0.04 = 0.09. (Key point: Curve type) 2. With R = 10000 m, how is the length and the BVC chainage obtained? A. L = N/R = 9e-6 m; BVC = 50+050 B. L = NR = 900 m; BVC = PVI chainage - L/2 = 49+600 C. L = NR = 900 m; BVC = PVI - L = 49+150 D. L = R/N; BVC = PVI + 450 Answer: B. The tangent points lie L/2 = 450 m either side of the PVI: BVC 49+600 and EVC 50+500. (Key point: Length from R) 3. What formula gives the RL at distance x from BVC on this summit curve? A. y = RL_BVC + g1 x + x^2/(2R) B. y = RL_PVI - x^2/(2R) C. y = RL_BVC + g1 x - x^2/(2R) D. y = RL_BVC - g1 x - x^2/R Answer: C. The parabola falls below the entering tangent by x^2/(2R) for a summit. (Key point: Elevation on the curve) 4. Why is x = 150 m used for the point 300 m before the PVI? A. Because the point is 300 m from BVC divided by 2 B. Because 150 m is half the radius per grade C. Because x is measured from the PVI D. Because x is measured from BVC, which is 450 m before the PVI: 450 - 300 = 150 Answer: D. BVC is 450 m before PVI, so the point is 150 m past BVC; RL = 478 + 7.5 - 1.125 = 484.375 m. 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