Setback on six lane highway, 550 m radius

Assumptions

Six-lane undivided carriageway with lane width $w=3.5$ m, $R=550$ m (road centre line), $L=350$ m. The driver follows the centre of the inner (kerb-side) lane, at $n=2(3.5)+1.75=8.75$ m from the centre line, so the path radius is $R_i=550-8.75=541.25$ m. Setback from the inner edge of the road = $m_c-w/2$.

Case 1: stopping sight distance S = 200 m ($S<L$)

$$\frac{\alpha}{2}=\frac{S}{2R_i}=0.18476\ \text{rad},\quad m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)=9.212\ \text{m}$$

Setback from the inner edge = 9.212 − 1.75 = 7.462 m.

Case 2: overtaking sight distance S = 400 m ($S>L$)

$$\frac{\alpha}{2}=\frac{L}{2R_i}=0.32333\ \text{rad}$$ $$m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)+\frac{S-L}{2}\sin\frac{\alpha}{2}=28.045+25\sin(18.525^\circ)=35.988\ \text{m}$$

Setback from the inner edge = 35.988 − 1.75 = 34.238 m.

Answer: about 7.46 m for SSD and 34.24 m for OSD, measured from the inner edge of the road.