Overtaking sight distance derivation

Overtaking Sight Distance (OSD) is the minimum distance open to the vision of the driver of a vehicle intending to overtake a slower vehicle ahead, so that the driver can overtake safely without any risk of collision with a vehicle coming from the opposite direction.

Factors affecting OSD

1. Speeds of the overtaking, overtaken and opposing vehicles.
2. Spacing between the vehicles (depends on speed).
3. Skill and reaction time of the driver.
4. Rate of acceleration of the overtaking vehicle.
5. Gradient of the road and the road surface condition.
6. Width of the road (two-lane/multi-lane) and the traffic volume.

Derivation (two-lane two-way road)

Let $A$ = overtaking vehicle, $B$ = overtaken vehicle (speed $v_b$ m/s), $C$ = oncoming vehicle; $v$ = design speed of $A$ and $C$ (m/s), $a$ = acceleration of $A$ (m/s²), $t$ = reaction time of the driver of $A$ (2 s).

 A  <-d1->|<-------- d2 -------->|<---- d3 ---->  C
 [A]..[B] |  [A]  ........ [B] [A]|   <- [C]
 start    |  overtaking path      | opposite car

$d_1$ - distance travelled during the reaction time (A travels at speed $v_b$ behind B):

$$d_1 = v_b\,t$$

$d_2$ - distance during the overtaking operation. A starts at spacing $s$ behind B and finishes at spacing $s$ ahead of B, so A gains $2s$ on B in time $T$. Relative to B, A starts with zero relative speed and accelerates at $a$:

$$\begin{aligned} 2s &= \tfrac12 a T^2 \ \Rightarrow\ T = \sqrt{\frac{4s}{a}} \\ d_2 &= v_b\,T + 2s \end{aligned}$$

where $s = 0.7v_b + 6$ (m, with $v_b$ in m/s).

$d_3$ - distance travelled by the oncoming vehicle C during the overtaking time $T$ (C runs at the design speed $v$):

$$d_3 = v\,T$$

Total

$$\boxed{OSD = d_1 + d_2 + d_3 = v_b t + (v_bT + 2s) + vT}$$

As per IRC, the speed of the overtaken vehicle is taken as $V_b = V - 16$ km/h (i.e. $v_b = v - 4.5$ m/s) when it is not known.

IRC gives about 470 m for $V = 80$ km/h on a two-way road (about 300 m for 60 km/h). On one-way roads $d_3$ is not needed, so $OSD = d_1 + d_2$.