Transition curve for 310 m radius

Given: $R = 310$ m, $V = 90$ km/h, two-lane NH in plain terrain, $c = 0.5$ m/s³, rate of introduction of superelevation 1 in 120, PI at $1+000$ m.

Reading: "angle of intersection 136°" is the interior angle between the straights, so the deflection angle $\Delta = 180^\circ - 136^\circ = 44^\circ$.

Assumptions: carriageway width $W = 7.0$ m, wheel base $l = 6$ m.

Step 1: Superelevation

$$e = \frac{V^2}{225R} = \frac{90^2}{225\times310} = 0.116 > 0.07$$

Adopt $e_{max} = 7\%$. Check: $f = \dfrac{90^2}{127\times310} - 0.07 = 0.136 < 0.15$, OK.

Step 2: Extra widening

$$W_e = \frac{nl^2}{2R} + \frac{V}{9.5\sqrt{R}} = \frac{2(6)^2}{2(310)} + \frac{90}{9.5\sqrt{310}} = 0.116 + 0.538 = 0.65\ \text{m}$$

Step 3: Length of transition curve

$v = 90/3.6 = 25$ m/s.

• Centrifugal acceleration: $L_s = \dfrac{v^3}{cR} = \dfrac{25^3}{0.5\times310} = 100.8$ m
• Superelevation rate (1 in 120, rotation about the inner edge): $L_s = N\,e\,(W+W_e) = 120\times0.07\times7.65 = 64.3$ m
• IRC empirical: $L_s = \dfrac{2.7V^2}{R} = \dfrac{2.7\times90^2}{310} = 70.5$ m

Adopt the largest, rounded up: $L_s = 105$ m.

Step 4: Shift and tangent length

$$\begin{aligned} s &= \frac{L_s^2}{24R} = \frac{105^2}{24(310)} = 1.48\ \text{m} \\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = (311.48)\tan22^\circ + 52.5 = 125.85 + 52.5 = 178.35\ \text{m} \end{aligned}$$

Step 5: Length of circular curve

$$\phi_s = \frac{L_s}{2R} = 9.70^\circ,\qquad L_c = R\,\Delta_{rad} - L_s = 310\times0.76794 - 105 = 133.06\ \text{m}$$

Step 6: Chainages

Answer: $L_s = 105$ m; the curve begins at chainage 821.65 m (0+821.65) and the circular curve ends at 1059.71 m (1+059.71).

If 136° were the deflection angle itself: TS would be at chainage 176.56 m and the end of the circular curve at 912.39 m.