Rankine's deflection angle data, 145°, 5°

Angle of intersection (interior) = 145°, so deflection angle $\Delta = 180^\circ-145^\circ = 35^\circ$. Degree of curve 5° for a 30 m arc.

Elements

$$\begin{aligned} R &= \frac{1718.87}{D} = \frac{1718.87}{5} = 343.77\ \text{m}\\ T &= R\tan\frac{\Delta}{2} = 343.77\tan 17.5^\circ = 108.39\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 210.00\ \text{m}\\ \text{Ch. of T}_1 &= 1580 - 108.39 = 1471.61\ \text{m}\\ \text{Ch. of T}_2 &= 1471.61 + 210.00 = 1681.61\ \text{m} \end{aligned}$$

Deflection angles

$\delta = 1718.87\,c/R$ minutes, rounded to the least count of 10″. Peg interval 30 m: first sub-chord $= 28.39$ m, last sub-chord $= 1.61$ m.

Check: final deflection $= \Delta/2 = 17^\circ30'00''$; the table gives 17°30′00″.

Answer: $R = 343.77$ m, $T = 108.39$ m, $L = 210.00$ m; T₁ = 1471.61 m, T₂ = 1681.61 m; deflection angles as tabulated.