--- title: "Vertical curve, -3% and 1.5% at chainage 2000" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-79a-vc-2000" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-79a-vc-2000 site: BCT Engineering --- # Vertical curve, -3% and 1.5% at chainage 2000 Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2079 Ashwin, 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-79a-vc-2000 - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question A -3% gradient meets a 1.5% gradient at a chainage of 2000 m and at the reduced level of 500 m if the design speed of the road is 100 km/hr, determine RL and chainage of the tangent points, lowest point on the curve. Assume height of head light is 0.75 m and take frictional coefficient f = 0.35 and α = 1°. ## Short answer SSD=182 m, headlight criterion gives L=190 m. BVC at chainage 1905 m, RL 502.85 m; EVC at 2095 m, RL 501.425 m; lowest point at 2031.67 m, RL 500.950 m. ## Answer A -3% grade meeting a +1.5% grade forms a **valley (sag) curve**. Its length is the larger of the headlight-sight-distance length and the comfort length (IRC / NRS 2070, parabolic curve). #### Data $g_1=-3\%,\ g_2=+1.5\%$, so $N=0.03+0.015=0.045$. $V=100$ km/h, $f=0.35$, reaction time $t=2.5$ s, headlight height $h_1=0.75$ m, beam angle $\alpha=1^\circ$. PVI at chainage 2000 m, RL 500.00 m. #### Stopping sight distance $$ SSD=0.278Vt+\frac{V^2}{254f}=0.278(100)(2.5)+\frac{100^2}{254(0.35)}=182.0\ \text{m} $$ #### Length of curve Headlight criterion (assuming $L>S$), with $2\tan1^\circ = 0.0349$: $$ L=\frac{NS^2}{2(h_1+S\tan\alpha)}=\frac{0.045(182.0)^2}{1.5+0.0349(182.0)}=189.8\ \text{m} $$ $L>S$, so the assumption holds. Comfort criterion (rate of change of centrifugal acceleration $C=0.6$ m/s³, $v=27.78$ m/s): $$ L=2\sqrt{\frac{Nv^3}{C}}=80.2\ \text{m} $$ Adopt **L = 190 m** (the headlight criterion governs). #### Tangent points Half length = 95 m. - BVC chainage = 2000 − 95 = **1905 m**; RL = 500 + 0.03(95) = **502.85 m** - EVC chainage = 2000 + 95 = **2095 m**; RL = 500 + 0.015(95) = **501.425 m** #### Lowest point Distance from BVC where the slope is zero: $$ x=\frac{|g_1|L}{N}=\frac{0.03(190)}{0.045}=126.67\ \text{m} $$ Chainage = 1905 + 126.67 = **2031.67 m**. RL of the lowest point (with $g_1 = -0.03$): $$ y=RL_{BVC}+g_1x+\frac{Nx^2}{2L}=502.85-0.03(126.67)+\frac{0.045(126.67)^2}{2(190)}=500.950\ \text{m} $$ **Answer:** L = 190 m; BVC at 1905 m (RL 502.85 m); EVC at 2095 m (RL 501.425 m); lowest point at chainage 2031.67 m, RL 500.950 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. -3% meeting +1.5% gives which curve and what is N? A. Summit, N = 0.045 B. Valley, N = 0.045 C. Valley, N = 0.015 D. Summit, N = 0.015 Answer: B. A downgrade meeting an upgrade is a valley curve with N = 0.03 + 0.015 = 0.045. (Key point: Type of curve) 2. SSD for V = 100 km/h, t = 2.5 s, f = 0.35 is about A. 182 m B. 90 m C. 250 m D. 70 m Answer: A. 0.278 x 100 x 2.5 = 69.5; 10000/(254 x 0.35) = 112.5; total about 182 m. (Key point: Stopping sight distance) 3. Headlight criterion gave 189.4 m and comfort 80.2 m. Which governs? A. Comfort, as it is shorter B. Headlight, as it is longer; adopt L = 190 m C. Neither D. The average Answer: B. The longer value satisfies both, so L = 190 m. (Key point: Governing criterion) 4. With L = 190 m and PVI at 2000 m, RL 500 m, BVC is at A. Chainage 1905 m, RL 502.85 m B. Chainage 1905 m, RL 497.15 m C. Chainage 2095 m, RL 501.425 m D. Chainage 1810 m, RL 505.7 m Answer: A. BVC = 2000 - 95 = 1905 m; RL = 500 + 0.03 x 95 = 502.85 m. (Key point: Tangent points) 5. Distance of lowest point from BVC is x = g1 L / N. Its value is A. 95 m B. 126.67 m C. 63.3 m D. 190 m Answer: B. x = 0.03 x 190/0.045 = 126.67 m, so chainage 2031.67 m. 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