Setback on six lane highway, 450 m curve

Assumptions

Six-lane undivided carriageway, lane width $w=3.5$ m (NRS 2070 / IRC), $R=450$ m to the road centre line, $L=220$ m. The driver keeps to the centre of the innermost (kerb-side) lane, which is $n=2(3.5)+1.75=8.75$ m from the centre line. The path radius is $R_i=450-8.75=441.25$ m. The clearance from the lane centre to the obstruction line is $m_c$; the setback from the inner edge of the lane is $m_c-w/2=m_c-1.75$ m.

(i) Stopping sight distance S = 100 m ($S<L$)

$$\frac{\alpha}{2}=\frac{S}{2R_i}=\frac{100}{2(441.25)}=0.11331\ \text{rad}$$ $$m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)=2.830\ \text{m}\ \Rightarrow\ \text{setback}=2.830-1.75=1.080\ \text{m}$$

(ii) Overtaking sight distance S = 310 m ($S>L$)

$$\frac{\alpha}{2}=\frac{L}{2R_i}=\frac{220}{2(441.25)}=0.24929\ \text{rad}$$ $$m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)+\frac{S-L}{2}\sin\frac{\alpha}{2}=13.640+45\sin(14.283^\circ)=24.742\ \text{m}$$

Setback from the lane edge = 24.742 − 1.75 = 22.992 m.

Answer: setback from the edge of the inner lane = 1.08 m for SSD = 100 m and 22.99 m for OSD = 310 m.