Setback on four lane carriageway, 400 m radius

Assumptions

Four-lane undivided carriageway, lane width $w=3.75$ m (so pavement = 15 m), $R=400$ m to the road centre line, $L=220$ m. The driver keeps to the centre of the inner (kerb-side) lane, which is $n=3.75+1.875=5.625$ m from the centre line. Path radius $R_i=400-5.625=394.375$ m. The setback from the inner edge of the road is $m_c-\dfrac{w}{2}=m_c-1.875$ m.

Case 1: stopping sight distance S = 152 m ($S<L$)

$$\frac{\alpha}{2}=\frac{S}{2R_i}=\frac{152}{2(394.375)}=0.19271\ \text{rad}$$ $$m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)=7.300\ \text{m}$$

Setback from inner edge $=7.300-1.875=\mathbf{5.43}$ m.

Case 2: overtaking sight distance S = 300 m ($S>L$)

$$\frac{\alpha}{2}=\frac{L}{2R_i}=\frac{220}{2(394.375)}=0.27892\ \text{rad}$$ $$m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)+\frac{S-L}{2}\sin\frac{\alpha}{2}=15.242+40\sin(15.981^\circ)=26.254\ \text{m}$$

Setback from inner edge $=26.254-1.875=\mathbf{24.38}$ m.

Answer: minimum setback from the inner edge = 5.43 m (SSD) and 24.38 m (OSD).