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Posted speed limit at head-on crash

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.

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?

Given: g1=+4%g_1 = +4\%, g2=−3.5%g_2 = -3.5\%, posted speed V=70V = 70 km/h, highest point at 120 m from the beginning of the vertical curve (BVC), eye height h1=1.08h_1 = 1.08 m, reaction time t=2.5t = 2.5 s, f=0.35f = 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, h2=0h_2 = 0. The sight distance to be checked is the stopping sight distance.

Step 1: Length of the vertical curve

N=g1−g2=0.04+0.035=0.075N = g_1 - g_2 = 0.04 + 0.035 = 0.075

Highest point from BVC: x=g1LNx = \dfrac{g_1 L}{N}

120=0.04 L0.075 ⇒ L=225 m120 = \frac{0.04\,L}{0.075}\ \Rightarrow\ L = 225\ \text{m}

Step 2: SSD required at the posted speed

SSD=0.278 V t+V2254f=0.278(70)(2.5)+702254×0.35=48.65+55.12=103.8 m\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<LS < L and h2=0h_2 = 0, H=1.08H = 1.08 m:

L=NS22H ⇒ S=2HLN=2(1.08)(225)0.075=80.5 mL = \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 m<L=225S = 80.5\ \text{m} < L = 225 m, so the assumption S<LS < 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 VV for which SSD = 80.5 m:

V288.9+0.695V−80.5=0 ⇒ V=59.2 km/h\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 8HL/N=161\sqrt{8HL/N} = 161 m, and two opposing cars need 2×SSD2\times SSD, so each SSD must not exceed 80.5 m, i.e. V≈59V \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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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