Setback on rural highway, 80 kmph

Ruling minimum radius

With $e_{max}=0.07$ and $f_{max}=0.15$ (IRC / NRS 2070):

$$R_{min}=\frac{V^2}{127(e+f)}=\frac{80^2}{127(0.22)}=229.1\ \text{m}$$

Adopt R = 230 m (to the road centre line).

Setback

Assumptions: two-lane road, lane width 3.5 m, so the driver's path is at the centre of the inner lane, $n=1.75$ m from the centre line and the path radius is $R_i=230-1.75=228.25$ m. Sight distance $S=200$ m is greater than $L=150$ m, so the angle is based on the curve length:

$$\frac{\alpha}{2}=\frac{L}{2R_i}=\frac{150}{2(228.25)}=0.32859\ \text{rad}=18.827^\circ$$ $$m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)+\frac{S-L}{2}\sin\frac{\alpha}{2}=12.212+25\sin(18.827^\circ)=20.279\ \text{m}$$

$m_c$ is measured from the centre of the inner lane. The setback from the edge of the inner lane (pavement edge) is

$$m=m_c-\frac{w}{2}=20.279-1.75=18.53\ \text{m}$$

(Measured from the road centre line this is 22.03 m.)

Answer: the obstruction must be kept at least 18.53 m from the edge of the inner lane.