Rankine's data, 8° curve, 187.62 m long chord

Degree of curve $D = 8^\circ$ for a 30 m arc, long chord $= 187.62$ m, chainage of IP $= 2552.00$ m, peg interval 30 m.

Step 1: Radius and deflection angle

$$R = \frac{1718.87}{8} = 214.86\ \text{m}$$ $$\sin\frac{\Delta}{2} = \frac{\text{LC}}{2R} = \frac{187.62}{2\times 214.86} = 0.43661 \ \Rightarrow\ \frac{\Delta}{2} = 25.8879^\circ,\ \Delta = 51.7759^\circ = 51°46′33″$$

Step 2: Tangent length, curve length, chainages

$$\begin{aligned} T &= R\tan\frac{\Delta}{2} = 214.86\tan 25.8879^\circ = 104.27\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 194.16\ \text{m}\\ \text{Ch. T}_1 &= 2552.00 - 104.27 = 2447.73\ \text{m}\\ \text{Ch. T}_2 &= 2447.73 + 194.16 = 2641.89\ \text{m} \end{aligned}$$

Step 3: Rankine's deflection angles

$\delta = 1718.87\,c/R$ minutes. First sub-chord $= 12.27$ m, last sub-chord $= 1.89$ m.

Check: final deflection $= \Delta/2 = 25.8879^\circ$ = 25°53′17″; computed 25°53′16″. Field check: the long chord T₁T₂ should measure 187.62 m.

Answer: $R = 214.86$ m, $\Delta = 51°46′33″$, $T = 104.27$ m, $L = 194.16$ m, T₁ = 2447.73 m, T₂ = 2641.89 m.