Highway circular curve, apex distance 15.876 m

Data: apex (external) distance $E = 15.876$ m, deflection angle $\Delta = 52^\circ10'$, chainage of IP = 0+455.50 (455.50 m), normal chord $C = 30$ m.

1. Radius from the apex distance

$$E = R\left(\sec\frac{\Delta}{2} - 1\right) \quad\Rightarrow\quad R = \frac{E}{\sec 26^\circ05' - 1} = \frac{15.876}{0.11339} = 140.01\ \text{m}$$

A suitable (rounded) radius of R = 140 m is adopted (it gives E = 15.875 m, practically the same as the given 15.876 m).

2. Curve elements

$$T = R\tan\frac{\Delta}{2} = 140\tan 26^\circ05' = 68.535\ \text{m}$$ $$L_c = \frac{\pi R\Delta}{180^\circ} = \frac{\pi\times 140\times 52.1667}{180} = 127.467\ \text{m}$$

• Chainage of T1 (start) $= 455.50 - 68.535 = 386.965$ m
• Chainage of T2 (end) $= 386.965 + 127.467 = 514.432$ m

3. Tangential (deflection) angles

For a chord $c$ the tangential angle between the tangent and the chord is

$$\delta = \frac{1718.9\,c}{R}\ \text{minutes}$$

The first chord is a sub-chord to the first 30 m peg; the last is a sub-chord to T2. Normal chord $= 30$ m, $\delta_{30} = 1718.9\times 30/140 = 6°08'20"$.

Check: the final cumulative angle 26°05'01" equals $\Delta/2 = 26^\circ05'$ ✓.

Answer: R = 140 m (computed 140.01 m), T = 68.53 m, $L_c$ = 127.47 m, T1 at 386.97 m and T2 at 514.43 m; the setting-out table is above.