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Setback on six lane road, 500 m radius

Short answer

Inner lane radius is 491.25 m and S=200 m exceeds L=100 m. Using the S>L formula with α/2=5.832°, minimum setback from the centre line is 16.37 m.

Calculate the minimum setback from centre line of road for a curve of radius 500 m for a six lane road to ensure safe visibility. The stopping sight distance is 200 m, lane width is 3.5 m and the curve length is 100 m and no extra width is to be provided.

Assumptions

Six-lane undivided carriageway, lane width w=3.5w=3.5 m, R=500R=500 m (road centre line), L=100L=100 m, SSD S=200S=200 m. No extra widening. The sight line is along the centre of the inner (kerb-side) lane, at n=2(3.5)+1.75=8.75n=2(3.5)+1.75=8.75 m from the centre line, so its radius is Ri=500−8.75=491.25R_i=500-8.75=491.25 m.

Check S against L

S=200S=200 m is greater than L=100L=100 m, so the sight line extends beyond the curve along the straight tangent and the angle is based on the curve length:

α2=L2Ri=1002(491.25)=0.10178 rad=5.832∘\frac{\alpha}{2}=\frac{L}{2R_i}=\frac{100}{2(491.25)}=0.10178\ \text{rad}=5.832^\circ

Setback from the centre line

m=R−(R−n)cos⁡α2+S−L2sin⁡α2m=R-(R-n)\cos\frac{\alpha}{2}+\frac{S-L}{2}\sin\frac{\alpha}{2} m=11.292+50sin⁡(5.832∘)=16.373 mm=11.292+50\sin(5.832^\circ)=16.373\ \text{m}

(Clearance from the centre of the inner lane is mc=7.623m_c=7.623 m; the road edge is 3×3.5=10.53\times3.5=10.5 m from the centre line.)

Answer: minimum setback from the centre line ≈\approx 16.37 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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