Curve R=300 m, deflection 36°, with checks

Given: $\Delta = 36^\circ$, $R = 300$ m, chainage of IP $= 1190$ m, peg interval 30 m.

Curve elements

$$\begin{aligned} T &= R\tan\frac{\Delta}{2} = 300\tan 18^\circ = 97.48\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 188.50\ \text{m}\\ \text{Long chord} &= 2R\sin\frac{\Delta}{2} = 2\times 300\times\sin 18^\circ = 185.41\ \text{m}\\ \text{Ch. of T}_1 &= 1190 - 97.48 = 1092.52\ \text{m}\\ \text{Ch. of T}_2 &= 1092.52 + 188.50 = 1281.02\ \text{m} \end{aligned}$$

Deflection angle table

First sub-chord $c_1 = 17.48$ m, last sub-chord $c_n = 21.02$ m, $\delta = 1718.87\,c/R$ minutes.

Checks during setting out

1. Angular check: the final cumulative deflection angle must equal $\Delta/2 = 18^\circ00'00''$. The table gives 18°00′00″.
2. Chord (linear) check: the last point, fixed by the last sub-chord from the previous peg, must fall on T₂, which is also located independently by measuring $T = 97.48$ m from the IP along the second tangent. Also, the measured long chord T₁T₂ should agree with $2R\sin(\Delta/2) = 185.41$ m.
3. Sum of chords: $c_1 + (n-2)\times 30 + c_n$ must equal $L = 188.50$ m.

Answer: $T = 97.48$ m, $L = 188.50$ m, long chord $= 185.41$ m, T₁ = 1092.52 m, T₂ = 1281.02 m; deflection angles as tabulated, final = 18°00′00″.