Derive tangential angle 1718.87 C/R

Derivation of tangential angle

Consider a circular curve of radius $R$ with centre $O$ and a chord $T_1P_1$ of length $C$ starting at the tangent point T₁ (T₁ is the PC).

   T1 ------------> tangent
     \ δ
      \
       \__ P1
        \ /
         O

For a short chord the arc and chord are nearly equal, so the angle at the centre is

$$\beta = \frac{C}{R}\ \text{radian}$$

The angle between a tangent and a chord is half the central angle subtended by the chord (tangent–chord theorem). So the tangential (deflection) angle is

$$\delta = \frac{\beta}{2} = \frac{C}{2R}\ \text{radian}$$

Converting radians to minutes (1 radian $= \dfrac{180}{\pi}\times 60 = 3437.747'$):

$$\delta = \frac{3437.747}{2}\cdot\frac{C}{R} = 1718.87\times\frac{C}{R}\ \text{minutes}$$

Deflection angles for laying out the curve

• The tangential angle $\delta_i$ for a chord $c_i$ is $1718.87\,c_i/R$ minutes. For the first sub-chord $c_1$, a full chord $c$ and the last sub-chord $c_n$ the angles are $\delta_1 = 1718.87c_1/R$, $\delta = 1718.87c/R$, $\delta_n = 1718.87c_n/R$.
• The deflection angle of a point is the angle at T₁ between the tangent and the chord to that point; it is the sum of the tangential angles up to that point: $\Delta_1=\delta_1$, $\Delta_2 = \delta_1 + \delta$, and so on.
• The theodolite is set at T₁ with zero on the IP. At each point, turn the telescope by the cumulative deflection angle and measure the chord from the previous point.
• Check: the last deflection angle is $\Delta/2$, where $\Delta$ is the deflection angle of the curve.