Highway curve through M 16 m from IP

Step 1: Deflection angle and position of M

The angle to the right between the tangents is the interior angle $\angle PQR = 140^\circ$, so the deflection angle is $\Delta = 180^\circ - 140^\circ = 40^\circ$.

Since $\angle PQM = 70^\circ = 140^\circ/2$, the point M lies on the line joining the IP (Q) to the centre of the curve (the bisector of the interior angle). M is on the curve, so $QM$ is the external (apex) distance:

        Q (IP)
        /|\
      P/ M \R      QM = E = 16 m
        \|/
         O

Step 2: Radius

$$E = R\left(\sec\frac{\Delta}{2} - 1\right)\ \Rightarrow\ R = \frac{E}{\sec 20^\circ - 1} = \frac{16}{1.06418 - 1} = 249.31\ \text{m}$$

Step 3: Tangent length, curve length, chainages

$$\begin{aligned} T &= R\tan 20^\circ = 90.74\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 174.05\ \text{m}\\ \text{Ch. of T}_1 &= 2250.00 - 90.74 = 2159.26\ \text{m}\\ \text{Ch. of T}_2 &= 2159.26 + 174.05 = 2333.31\ \text{m} \end{aligned}$$

Step 4: First three deflection angles

The first sub-chord runs from T₁ to the next 30 m peg, $c_1 = 0.74$ m, followed by full chords of 30 m. $\delta = 1718.87\,c/R$ minutes.

Answer: $R = 249.31$ m; the first three deflection angles are 0°05′06″, 3°31′57″ and 6°58′47″.