Setback with overtaking, deflection 70 degrees

Given: two-lane road, $e = 6\%$, $f = 0.15$, $V = 75$ km/h, SSD $= 115$ m, deflection angle $\Delta = 70^\circ$.

Assumptions: lane width $w = 3.5$ m, so the line of sight runs along the centre of the inner lane at $d = 1.75$ m from the road centre line. Since there is not enough space for full overtaking sight distance, the intermediate sight distance is provided: $ISD = 2\times SSD$.

Step 1: Radius of the curve (centre line)

$$R = \frac{V^2}{127\,(e+f)} = \frac{75^2}{127\,(0.06+0.15)} = \frac{5625}{26.67} = 210.9\ \text{m}$$

Step 2: Sight distance and length of curve

$$S = ISD = 2\times115 = 230\ \text{m}$$

Length of curve $L_c = R\Delta = 210.9\times\dfrac{70\pi}{180} = 257.7$ m. Since $S = 230$ m $< L_c = 257.7$ m, the sight line lies fully within the curve (Case $S < L_c$).

Step 3: Setback

$$\alpha = \frac{S}{R-d} = \frac{230}{210.9 - 1.75} = 1.0996\ \text{rad} = 63.0^\circ$$ $$\begin{aligned} m &= R - (R-d)\cos\frac{\alpha}{2} \\ &= 210.91 - 209.16\cos(31.5^\circ) \\ &= 210.91 - 178.34 = 32.58\ \text{m} \end{aligned}$$

Answer: setback distance from the centre line of the road $m \approx 32.6$ m (about 33 m) for intermediate sight distance of 230 m.

For comparison, SSD only (115 m) would need $m \approx 9.6$ m.