Composite curve by deflection angle method

A composite curve is a circular curve with a transition (spiral) curve at each end. The deflection angle method sets out the transition curves from T1 and T2 and the central circular curve from the junction points.

                 IP
                /  \
               /    \
      T1 ~~~S1 ______ S2~~~ T2
          (spiral) (arc) (spiral)
         |<-L->|<-Lc->|<-L->|

Elements to compute

$R$ = radius of the circular arc, $L$ = length of transition (from $L = V^3/(\alpha R)$), $\Delta$ = deflection angle.

• Shift: $s = \dfrac{L^2}{24R}$
• Spiral angle: $\phi_s = \dfrac{L}{2R}$ rad
• Tangent length: $T_s = (R + s)\tan\dfrac{\Delta}{2} + \dfrac{L}{2}$
• Central angle of the circular arc: $\Delta_c = \Delta - 2\phi_s$; length $L_c = R\Delta_c$ (in radians)
• Chainages: T1 = IP − $T_s$; S1 = T1 + L; S2 = S1 + $L_c$; T2 = S2 + L

Setting out

1. First transition (from T1). Set the theodolite at T1 with zero reading on the IP. For a peg at distance $l$ along the curve (cubic spiral $y = x^3/6RL$):

$$\delta = \frac{l^2}{6RL}\ \text{rad} = \frac{573\,l^2}{RL}\ \text{minutes}$$

The pegs are set out by turning angle $\delta$ and measuring the chord (taken equal to the arc). The last peg at $l = L$ is S1 with $\delta_L = \phi_s/3$. Check the position of S1 from the data.

2. Circular curve (from S1). Shift the instrument to S1. Take a back-sight on T1. The angle between the chord S1T1 and the common tangent at S1 is $\phi_s - \phi_s/3 = \dfrac{2\phi_s}{3}$. So, after the back-sight, plunge the telescope and turn it through $\dfrac{2\phi_s}{3}$ to get the direction of the tangent at S1, and set the horizontal circle to zero there. Then set out the arc by Rankine's deflection angles from this tangent:

$$\delta = \frac{1718.9\,c}{R}\ \text{minutes}$$

for a chord $c$ (first sub-chord, full chords, last sub-chord). The cumulative angle at S2 must equal $\Delta_c/2$, which checks the work.

3. Second transition (from T2). Set up at T2, back-sight the IP (the tangent), and set out pegs backwards from T2 towards S2 using the same deflection angles $\delta = 573\,l^2/(RL)$ minutes, where $l$ is the distance of the peg from T2. The pegs should meet the circular curve at S2.

Check: The pegs of the circular curve from S1 and the pegs of the second transition from T2 must meet at S2.

The method needs only a theodolite and a tape, but it is slow because of the shifting of the instrument at the junction points.