Minimum sight distance to avoid head-on collision

For head-on approach, both drivers must stop in the total distance available, so the required sight distance is the sum of the stopping distances of the two vehicles. On a 10% grade one vehicle goes uphill and the other downhill.

Data

$V=60$ km/h each, $t=2.5$ s, $f=0.40$, brake efficiency 50%, so effective $f=0.40\times0.5=0.20$. Gradient $n=0.10$.

• Lag distance (each): $0.278Vt=0.278(60)(2.5)=41.7$ m

Vehicle going uphill (grade helps braking)

$$SD_1=0.278Vt+\frac{V^2}{254(f+n)}=41.7+\frac{3600}{254(0.30)}=88.9\ \text{m}$$

Vehicle going downhill (grade opposes braking)

$$SD_2=0.278Vt+\frac{V^2}{254(f-n)}=41.7+\frac{3600}{254(0.10)}=183.4\ \text{m}$$

Minimum sight distance

$$SD=SD_1+SD_2=88.9+183.4=272.4\ \text{m}$$

Answer: minimum sight distance to avoid head-on collision $\approx$ 272 m. (On level ground it would be $2\times112.6=225.1$ m.)