--- title: "Vertical curve +4% and -3%, through given point" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-70m-vc-4-3" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-70m-vc-4-3 site: BCT Engineering --- # Vertical curve +4% and -3%, through given point Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2070 Magh, 8 marks - Video: [Transportation-Elevation at a Certain Point on a Vertical Curve](https://www.youtube.com/watch?v=P_xoN5BHBo0) — Civil Engineering Academy - 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-70m-vc-4-3 - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question An ascending 4% gradient meets with 3% descending gradient. Design the length of vertical curve to meet the visibility requirement. Calculate the RL of beginning end and highest points on the curve if the curve should pass through the point having elevation 120.10 m located at 60 m right from PVI. Assume design speed as 80 kmph, brake efficiency as 90% and coefficient of friction as 0.35. ## Short answer Gradient difference N = 0.07, SSD = 135.6 m. Assuming L > S, length L = NS^2/4.4 = 295 m. From point 60 m right of PVI (elevation 120.10 m), RL of PVI = 122.808 m. RL of BVC = 116.91 m, highest point = 120.28 m, RL of EVC = 118.38 m. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A +4% grade meets a -3% grade. What type of vertical curve is it and what is N? A. Summit curve, N = 0.07 B. Valley curve, N = 0.07 C. Summit curve, N = 0.01 D. Valley curve, N = 0.01 Answer: A. An ascending grade followed by a descending grade makes a summit curve; N is the algebraic difference 0.04 + 0.03 = 0.07. (Key point: Curve type) 2. With 90% brake efficiency and f = 0.35, what friction value is used in the braking distance term of SSD? A. 0.35 B. 0.389 C. 0.315 D. 0.45 Answer: C. Brake efficiency reduces friction: 0.35 x 0.90 = 0.315. (Key point: Effective friction) 3. The 120.10 m point is 60 m on the EVC side of the PVI. How is RL of the PVI found from it? A. Subtract the curve offset from 120.10 only B. Add the tangent drop (0.03 x 60) and the curve offset N x'^2/(2L) to 120.10 C. Add 0.04 x 60 and ignore the offset D. Subtract both the tangent drop and the offset from 120.10 Answer: B. The curve lies below the tangent, and the tangent itself is 1.80 m below the PVI at 60 m, so RL_PVI = 120.10 + 1.80 + 0.908. (Key point: Offset below tangent) 4. Where is the highest point of the summit curve measured from BVC? A. x = L/2 = 147.5 m B. x = g2 L / N = 126.4 m C. x = N L / 8 = 2.58 m D. x = g1 L / N = 168.57 m Answer: D. The highest point is where the grade is zero: g1 - (N/L)x = 0, giving x = g1 L / N. 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