Rankine's table, IP 2+450 km, 30°, 3° curve

Chainage of IP $= 2+450$ km $= 2450$ m, $\Delta = 30^\circ$, degree of curve $D = 3^\circ$ for a fixed arc of 25 m, peg interval 30 m.

Elements

$$\begin{aligned} R &= \frac{180\,C}{\pi D} = \frac{180\times 25}{\pi\times 3} = 477.46\ \text{m}\\ T &= R\tan 15^\circ = 127.94\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 250.00\ \text{m}\\ \text{Ch. T}_1 &= 2450 - 127.94 = 2322.06\ \text{m}\\ \text{Ch. T}_2 &= 2322.06 + 250.00 = 2572.06\ \text{m} \end{aligned}$$

Setting-out table (theodolite at T₁)

$\delta = 1718.87\,c/R$ minutes:

Check: final deflection $= \Delta/2 = 15^\circ00'00''$ (computed 15°00′00″).

Answer: $R = 477.46$ m, $T = 127.94$ m, $L = 250.00$ m, T₁ = 2322.06 m, T₂ = 2572.06 m; deflection angles as tabulated.