Simple curve data, roads intersecting at 150°

Interior angle at A = 150°, so the roads deflect by $\Delta = 180^\circ - 150^\circ = 30^\circ$. A 4° curve with a 30 m chain means the degree of curve is defined by a 30 m arc/chord.

Radius

$$R = \frac{1718.87}{D} = \frac{1718.87}{4} = 429.72\ \text{m}$$

Curve elements

$$\begin{aligned} T &= R\tan\frac{\Delta}{2} = 429.72\tan 15^\circ = 115.14\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = \frac{\pi\times 429.72\times 30}{180} = 225.00\ \text{m}\\ \text{Long chord} &= 2R\sin\frac{\Delta}{2} = 2\times 429.72\sin 15^\circ = 222.44\ \text{m}\\ \text{Apex distance} &= R\left(\sec\frac{\Delta}{2} - 1\right) = 15.16\ \text{m}\\ \text{Mid-ordinate} &= R\left(1-\cos\frac{\Delta}{2}\right) = 14.64\ \text{m} \end{aligned}$$

Chainages

Chainage of A $= 138\times 30 + 20.3 = 4160.30$ m.

• Chainage of T₁ (start) $= 4160.30 - 115.14 = 4045.16$ m
• Chainage of T₂ (end) $= 4045.16 + 225.00 = 4270.16$ m

Deflection angles (30 m chain, Rankine)

The first sub-chord is the distance from T₁ to the next whole 30 m station, then full chords, and a last sub-chord to T₂.

The last cumulative deflection equals $\Delta/2 = 15^\circ$ (check).

Answer: $\Delta = 30^\circ$, $R = 429.72$ m, $T = 115.14$ m, apex distance $= 15.16$ m, mid-ordinate $= 14.64$ m, $L = 225.00$ m, long chord $= 222.44$ m.