--- title: "Posted speed limit at head-on crash" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-79j-posted-speed" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-79j-posted-speed site: BCT Engineering --- # Posted speed limit at head-on crash Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2079 Jestha, 8 marks; 2078 Chaitra, 8 marks (asked 2 times) - Video: [Overtaking Sight Distance (OSD) Solved Numerical – Head-on Collision Avoidance | Highway Engineering](https://www.youtube.com/watch?v=NLFDSpEp9QA) — Civil Engineering Adhyayan - 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-79j-posted-speed - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question Two approaching cars were caught on head-on collision at a vertical summit curve connecting 4% ascending gradient with 3.5% descending gradient. The posted speed on this road is 70 km/hr. The highway engineer checked the geometrics of the road and found that the highest point of the vertical curve lies at the distance of 120 m from the beginning of the vertical curve. Check if the crash is due to fault in posted speed sign. Height of car driver's eye = 1.08 m, reaction time = 2.5 sec, coefficient of longitudinal friction = 0.35. What should be the posted speed if there is fault in posted speed limit sign? ## Short answer Available sight distance at the summit curve is 80.5 m, but required SSD at 70 km/h is 103.8 m, so the posted speed is unsafe. The crash is due to a faulty sign; the safe posted speed should be 55 km/h. ## Answer **Given:** $g_1 = +4\%$, $g_2 = -3.5\%$, posted speed $V = 70$ km/h, highest point at 120 m from the beginning of the vertical curve (BVC), eye height $h_1 = 1.08$ m, reaction time $t = 2.5$ s, $f = 0.35$. **Assumption:** only the driver's eye height is given, so the object (the other car or an obstruction) is taken at road level, $h_2 = 0$. The sight distance to be checked is the stopping sight distance. #### Step 1: Length of the vertical curve $$ N = g_1 - g_2 = 0.04 + 0.035 = 0.075 $$ Highest point from BVC: $x = \dfrac{g_1 L}{N}$ $$ 120 = \frac{0.04\,L}{0.075}\ \Rightarrow\ L = 225\ \text{m} $$ #### Step 2: SSD required at the posted speed $$ \begin{aligned} SSD &= 0.278\,V\,t + \frac{V^2}{254 f} \\ &= 0.278(70)(2.5) + \frac{70^2}{254\times0.35} \\ &= 48.65 + 55.12 = 103.8\ \text{m} \end{aligned} $$ #### Step 3: Sight distance available on the curve For a summit curve with $S < L$ and $h_2 = 0$, $H = 1.08$ m: $$ L = \frac{N S^2}{2H}\ \Rightarrow\ S = \sqrt{\frac{2HL}{N}} = \sqrt{\frac{2(1.08)(225)}{0.075}} = 80.5\ \text{m} $$ $S = 80.5\ \text{m} < L = 225$ m, so the assumption $S < L$ is valid. #### Step 4: Check Available sight distance 80.5 m $<$ required SSD 103.8 m. **The sight distance is not sufficient at 70 km/h, so the posted speed sign is faulty.** #### Step 5: Safe posted speed Find $V$ for which SSD = 80.5 m: $$ \frac{V^2}{88.9} + 0.695V - 80.5 = 0\ \Rightarrow\ V = 59.2\ \text{km/h} $$ Check: at 60 km/h, SSD = 82.2 m (slightly greater than 80.5 m); at 55 km/h, SSD = 72.3 m (< 80.5 m). **Answer: The posted 70 km/h is unsafe (SSD 103.8 m > available 80.5 m). The posted speed should be 55 km/h** (the calculated limit is about 59 km/h, rounded down to a practical value of 55 km/h). *Note:* the head-on reading gives the same result. With both drivers' eyes at 1.08 m, the available sight distance is $\sqrt{8HL/N} = 161$ m, and two opposing cars need $2\times SSD$, so each SSD must not exceed 80.5 m, i.e. $V \approx 59$ km/h. If the IRC object height of 0.15 m is used instead, the available distance becomes 110.5 m and 70 km/h would be acceptable (maximum 72.9 km/h); the answer above uses the data as given. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is L found from the highest point at 120 m from BVC? A. L = g1 N / x B. x = g1 L / N, so L = 120 x 0.075/0.04 = 225 m C. L = 2 x 120 D. L = 120 / N Answer: B. The highest point lies at x = g1 L/N, so L = 120 x 0.075/0.04 = 225 m. (Key point: Length of curve) 2. SSD at 70 km/h with t = 2.5 s and f = 0.35 is about A. 80.5 m B. 55 m C. 103.8 m D. 150 m Answer: C. SSD = 0.278(70)(2.5) + 70^2/(254 x 0.35) = 48.65 + 55.12 = 103.8 m. (Key point: SSD required) 3. For S < L and object height 0, the available sight distance is S = sqrt(2HL/N) = A. 80.5 m B. 103.8 m C. 110.5 m D. 225 m Answer: A. S = sqrt(2 x 1.08 x 225 / 0.075) = 80.5 m. (Key point: Available sight distance) 4. What does comparing 80.5 m available with 103.8 m required show? A. The sign is correct B. The sight distance is insufficient, so the posted speed is faulty C. The curve is too long D. The curve is a valley curve Answer: B. Available is less than required, so 70 km/h is unsafe; a lower speed of about 55-59 km/h is needed. 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