Transition curve 625 m radius, 90 kmph

Reading of the data

The angle of intersection 130°24′ is taken as the interior angle between the straights, so the deflection angle is $\Delta=180^\circ-130^\circ24'=49^\circ36'=49.6^\circ$. (With $\Delta=130.4^\circ$ the tangent length would exceed the chainage of the PI and give a negative starting chainage, so this reading is the consistent one.)

(a) Length of transition curve

$V=90/3.6=25$ m/s, $C=0.25$ m/s³:

$$L_s=\frac{V^3}{CR}=\frac{25^3}{0.25(625)}=100.0\ \text{m}$$

Adopt $L_s=100$ m.

(b) Chainages of beginning and end of the curve

• Shift: $s=\dfrac{L_s^2}{24R}=\dfrac{100^2}{24(625)}=0.667$ m
• Spiral angle: $\theta_s=\dfrac{L_s}{2R}=0.0800$ rad $=4.584^\circ$
• Central angle of circular arc: $\Delta_c=\Delta-2\theta_s=49.6^\circ-9.167^\circ=40.433^\circ$
• Circular arc length: $L_c=R\Delta_c=441.05$ m
• Total length of combined curve: $L=L_c+2L_s=641.05$ m
• Tangent length from PI:

$$T_s=(R+s)\tan\frac{\Delta}{2}+\frac{L_s}{2}=(625.667)\tan(24.80^\circ)+50=339.10\ \text{m}$$

(c) Chainages (PI at 1092.500 m)

• Beginning of curve (TS) = 1092.500 − 339.10 = 753.40 m
• End of curve (ST) = 753.40 + 641.05 = 1394.45 m

Answer: $L_s=100$ m; curve begins at 753.40 m and ends at 1394.45 m.