Stopping sight distance and derivation

Stopping Sight Distance (SSD) is the minimum distance ahead along the road that a driver must be able to see an object (height 0.15 m) on the carriageway so that the vehicle running at the design speed can be stopped safely before reaching it. It is the sum of the lag distance and the braking distance.

Factors affecting SSD

1. Speed of the vehicle (increases SSD as $V^2$).
2. Total reaction time of the driver (PIEV, 2.5 s as per IRC/NRS).
3. Coefficient of longitudinal friction $f$ between tyre and pavement (0.35-0.40; depends on surface, tyre and weather).
4. Gradient of the road (down-grade increases SSD).
5. Brake efficiency and tyre condition; weight of the vehicle.
6. Visibility conditions such as night, fog, and the height of eye and object.

Derivation (level road)

Let $v$ = speed (m/s), $t$ = reaction time (s), $f$ = coefficient of friction, $l$ = lag distance, $l_b$ = braking distance.

Lag distance (vehicle moves at constant speed during reaction time):

$$l = v\,t$$

Braking distance: work done by friction = loss of kinetic energy

$$\begin{aligned} f\,W\,l_b &= \frac{1}{2}\frac{W}{g}v^2 \\ l_b &= \frac{v^2}{2gf} \end{aligned}$$

Therefore

$$SSD = v\,t + \frac{v^2}{2gf}$$

In terms of $V$ (km/h): $v = 0.278V$

$$\boxed{SSD = 0.278\,V\,t + \frac{V^2}{254\,f}}$$

For a gradient $n\%$ the braking term becomes $\dfrac{V^2}{254(f \pm 0.01n)}$ (+ upgrade, - downgrade).

Example: $V = 60$ km/h, $t = 2.5$ s, $f = 0.35$: $SSD = 41.7 + 40.5 = 82.2$ m.