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Setback on 400 m curve, 1.95 m lane distance

Short answer

For SSD = 90 m (< L), setback m = 4.49 m from road centre line. For OSD = 300 m (> L), setback = 26.87 m.

There is a horizontal curve with radius of 400 m and length of 200 m on the highway. Compute the set-back distance required from the centre line on the inner side of the curve so as to provide for (i) stopping sight distance of 90 m (ii) safe overtaking sight distance of 300 m. The distance between the centre line of the road and inner lane is 1.95 m.

Data

R=400R=400 m (centre line), L=200L=200 m, n=1.95n=1.95 m (centre line of road to centre of inner lane). The sight line runs along the centre of the inner lane, radius R−n=398.05R-n=398.05 m. The setback mm is measured from the road centre line.

(i) SSD = 90 m (S<LS<L)

α2=S2(R−n)=902(398.05)=0.11305 rad=6.477∘\frac{\alpha}{2}=\frac{S}{2(R-n)}=\frac{90}{2(398.05)}=0.11305\ \text{rad}=6.477^\circ m=R−(R−n)cos⁡α2=400−398.05cos⁡(6.477∘)=4.491 mm=R-(R-n)\cos\frac{\alpha}{2}=400-398.05\cos(6.477^\circ)=4.491\ \text{m}

(ii) OSD = 300 m (S>LS>L)

α2=L2(R−n)=2002(398.05)=0.25122 rad=14.394∘\frac{\alpha}{2}=\frac{L}{2(R-n)}=\frac{200}{2(398.05)}=0.25122\ \text{rad}=14.394^\circ m=R−(R−n)cos⁡α2+S−L2sin⁡α2=14.445+50sin⁡(14.394∘)=26.875 mm=R-(R-n)\cos\frac{\alpha}{2}+\frac{S-L}{2}\sin\frac{\alpha}{2}=14.445+50\sin(14.394^\circ)=26.875\ \text{m}

Answer: setback from the centre line = 4.49 m (SSD = 90 m) and 26.87 m (OSD = 300 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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