Degree of curve, tangential angle formula

Degree of curve

The degree of curve $D$ is the central angle subtended by a standard chord (arc) of fixed length $C$ (30 m in the metric system, 100 ft in the foot system). A sharper curve has a larger degree and a smaller radius.

              O
             /|\
            / D \
           /  |  \
          A---+---B   chord/arc C = 30 m

Relation to the radius: central angle in radians $= C/R$, so

$$D = \frac{180^\circ}{\pi}\cdot\frac{C}{R}\quad\Rightarrow\quad R = \frac{180\,C}{\pi D} = \frac{1718.87}{D}\ \ (C = 30\ \text{m})$$

Derivation of tangential angle $\alpha = \dfrac{90^\circ C}{\pi R}$

   T1 ---------------- IP
    \ a
     \___ P1
      \  \  C
       \  \
        O (centre)

Let T₁P₁ be a chord of length $C$ at the start of the curve. The tangent at T₁ is perpendicular to the radius $OT_1$. The chord is short, so it is nearly equal to its arc, and the central angle subtended by it is

$$\angle T_1OP_1 = \frac{C}{R}\ \text{radians} = \frac{180}{\pi}\cdot\frac{C}{R}\ \text{degrees}$$

The angle between a tangent and a chord equals half the angle subtended by the chord at the centre (tangent–chord theorem). Hence the tangential angle

$$\alpha = \frac{1}{2}\cdot\frac{180^\circ C}{\pi R} = \frac{90^\circ C}{\pi R}$$

In minutes, $\alpha = 1718.87\,C/R$.

Derivation of the deflection angle $\Delta_n$

Deflection angle of the $n$th point is the angle at T₁ between the initial tangent and the chord T₁P$_n$. For successive chords with central angles $\beta_1, \beta_2, \dots, \beta_n$, the arc T₁P$_n$ subtends $\beta_1+\beta_2+\dots+\beta_n$ at the centre. By the same theorem

$$\Delta_n = \frac{1}{2}(\beta_1+\beta_2+\dots+\beta_n) = \delta_1+\delta_2+\delta_3+\dots+\delta_n$$

where $\delta_i = \beta_i/2 = 1718.87\,c_i/R$ minutes. Thus the deflection angle of any point is the sum of the tangential angles of all chords up to it. For the whole curve $\Delta_{last} = \Delta/2$, which is used as the check in the field.