--- title: "Vertical curve for overtaking sight distance" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-81c-vc-overtaking" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-81c-vc-overtaking site: BCT Engineering --- # Vertical curve for overtaking sight distance Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2081 Chaitra, 8 marks - Video: [Overtaking sight distance problems || Problems solved on overtaking and overtaken vehicles...](https://www.youtube.com/watch?v=SRVXJuapv6M) — Engineering Podium - 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-81c-vc-overtaking - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question An ascending gradient of 1.2% meets a descending gradient of 1.08% on a two lane two-way road. Determine the length of vertical curve to provide overtaking sight distance for a design speed of 60 kmph. Assume overtaking acceleration as 2.5 kmphps. Locate the highest point and end of curve, calculate their elevations if the elevation of BVC is 334.792 m. ## Short answer For grades +1.2% and -1.08% (N=0.0228) at 60 km/h with overtaking acceleration 2.5 kmphps, OSD is 318.1 m. Summit curve length is 215.1 m (using S>L formula L = 2S - 9.6/N). Highest point is 113.2 m from BVC at RL 335.47 m. End of curve is at RL 334.92 m (215.1 m from BVC at RL 334.792 m). ## Answer **Given:** $g_1 = +1.2\%$, $g_2 = -1.08\%$, two-lane two-way road, $V = 60$ km/h, acceleration $A = 2.5$ km/h/s, BVC elevation $= 334.792$ m. $$ N = 0.012 + 0.0108 = 0.0228 $$ **Assumptions (IRC):** reaction time $t = 2$ s; speed of the overtaken vehicle $v_b = V - 16 = 44$ km/h $= 12.22$ m/s; spacing $s = 0.7v_b + 6$ with $v_b$ in m/s (equivalently $0.2V_b + 6$ with $V_b$ in km/h); eye height and object height both 1.2 m for overtaking. $v = 60/3.6 = 16.67$ m/s. #### Step 1: Overtaking sight distance $$ \begin{aligned} d_1 &= v_b\,t = 12.22\times2 = 24.4\ \text{m} \\ s &= 0.7(12.22) + 6 = 14.56\ \text{m} \\ T &= \sqrt{\frac{14.4\,s}{A}} = \sqrt{\frac{14.4\times14.56}{2.5}} = 9.16\ \text{s} \\ d_2 &= v_bT + 2s = 12.22(9.16) + 2(14.56) = 141.0\ \text{m} \\ d_3 &= vT = 16.67\times9.16 = 152.6\ \text{m} \end{aligned} $$ (The factor 14.4 is $4\times3.6$, because $A$ is in km/h/s and $s$ in metres.) $$ OSD = 24.4 + 141.0 + 152.6 = 318.1\ \text{m} $$ #### Step 2: Length of summit curve Try $S < L$: $L = \dfrac{NS^2}{9.6} = \dfrac{0.0228(318.1)^2}{9.6} = 240.3$ m, which is less than $S$, so the assumption fails. Use $S > L$: $$ L = 2S - \frac{9.6}{N} = 2(318.1) - \frac{9.6}{0.0228} = 636.1 - 421.1 = 215.1\ \text{m} $$ $S = 318.1 > L = 215.1$ m, consistent. **$L = 215.1$ m.** #### Step 3: Highest point $$ x = \frac{g_1L}{N} = \frac{0.012\times215.1}{0.0228} = 113.2\ \text{m from BVC} $$ $$ RL_{high} = 334.792 + 0.012(113.2) - \frac{0.0228\,(113.2)^2}{2\times215.1} = 334.792 + 1.358 - 0.679 = 335.47\ \text{m} $$ #### Step 4: End of curve (EVC) $$ RL_{EVC} = 334.792 + \frac{(0.012-0.0108)(215.1)}{2} = 334.792 + 0.129 = 334.92\ \text{m} $$ The EVC is 215.1 m from the BVC. **Answer:** OSD $= 318.1$ m; $L = 215.1$ m; highest point 113.2 m from BVC at RL $335.47$ m; end of curve at RL $334.92$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Overtaking sight distance is the sum of A. d1 (reaction), d2 (overtaking), d3 (opposing vehicle distance) B. braking distance and lag distance C. d1 and d2 only D. three times the stopping distance Answer: A. OSD = d1 + d2 + d3; here 24.4 + 141.0 + 152.6 = 318.1 m. (Key point: Components of OSD) 2. The time of overtaking T = sqrt(14.4 s / A) is used because A. A is in km/h/s and s in metres B. A is in m/s^2 and s in km C. T is in minutes D. s is a time Answer: A. 14.4 = 4 x 3.6 accounts for A in km/h/s with s in metres, giving T = 9.16 s. (Key point: Overtaking time) 3. The trial L = N S^2/9.6 = 240.3 m is less than S = 318.1 m. What does this mean? A. The assumption S < L fails, so the S > L formula is used B. The answer 240.3 m is final C. The curve is a valley curve D. The speed is too low Answer: A. A result smaller than S contradicts S < L, so use L = 2S - 9.6/N = 215.1 m. (Key point: Choosing S < L or S > L) 4. How far from BVC is the highest point? A. About 215 m B. About 107 m C. About 113 m D. About 226 m Answer: C. x = g1 L/N = 0.012 x 215.1/0.0228 = 113.2 m. (Key point: Highest point) ## Related in Transportation Engineering I - [Overtaking sight distance and zone, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-76b-osd-60) — Transportation Engineering I, chapter 3 - [Vertical curve with intermediate sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81a-vc-intermediate) — Transportation Engineering I, chapter 3 - [Sight distance with obstruction on curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78b-sd-obstruction) — Transportation Engineering I, chapter 3 - [Overtaking sight distance and zone numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-62j-osd-numeric) — Transportation Engineering I, chapter 3 - [Vertical curve +2% and +5.5%, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79c-vc-2-55) — Transportation Engineering I, chapter 3 - [Stopping sight distance at 50 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-66m-ssd-50) — Transportation Engineering I, chapter 3 ## Related in other subjects - [Transition curve, road 8 m, R=300 m, 60°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-300m-60-70kmph) — Surveying II, chapter 6, asked 2× - [Acceleration for 25 kmph schedule speed, 800 m stations](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aacceleration-25kmph-800m-stations) — Utilization of Electrical Energy, chapter 4 - [Quadrilateral speed-time curve on down gradient](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aquadrilateral-curve-down-gradient-energy) — Utilization of Electrical Energy, chapter 4 - [Transition curve road 8 m, R=330 m, 60°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-330m-deflection-60) — Surveying II, chapter 6, asked 2×