Transition curve 400 m radius, 85 kmph

Given: $R = 400$ m, $V = 85$ km/h, $c = 0.52$ m/s³, intersection angle (interior) $= 125^\circ 25'$, PI chainage $= 1075.5$ m.

Reading: the "angle of intersection" is the interior angle between the straights, so the deflection angle $\Delta = 180^\circ - 125^\circ25' = 54^\circ35' = 54.583^\circ$.

(i) Length of transition curve

$v = 85/3.6 = 23.61$ m/s.

$$L_s = \frac{v^3}{cR} = \frac{(23.61)^3}{0.52\times400} = \frac{13162.6}{208} = 63.3\ \text{m}$$

Check with the IRC empirical formula (plain/rolling): $2.7V^2/R = 2.7(85)^2/400 = 48.8$ m $<63.3$ m. The rate of superelevation criterion is not requested (no width data). Adopt

$L_s = 65$ m.

(ii) Shift

$$s = \frac{L_s^2}{24R} = \frac{65^2}{24\times400} = \frac{4225}{9600} = 0.44\ \text{m}$$

(iii) Chainages

Spiral angle $\phi_s = \dfrac{L_s}{2R} = 0.0813$ rad $= 4.66^\circ$.

$$\begin{aligned} T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} \\ &= (400.44)\tan(27.292^\circ) + 32.5 = 206.6 + 32.5 = 239.1\ \text{m} \end{aligned}$$

Length of circular curve:

$$L_c = R\left(\Delta - 2\phi_s\right) = 400\,(0.95269 - 0.16250) = 316.1\ \text{m}$$

Answer: $L_s = 65$ m; shift $= 0.44$ m; curve begins at chainage 836.4 m and ends at 1282.5 m (total length $316.1 + 130 = 446.1$ m).

If the 125°25' were taken as the deflection angle instead, $T_s = 808.6$ m, the curve would begin at 266.9 m and end at 1207.5 m.