Composite curve R=400 m, Ls=90 m, IP 1+400

Given: $\Delta = 40^\circ$, $R = 400$ m, $L_s = 90$ m, chainage of IP $= 1+400$ km $= 1400$ m, peg interval 30 m on the transitions and 20 m on the circular curve.

Curve elements

$$\begin{aligned} s &= \frac{L_s^2}{24R} = \frac{90^2}{24\times 400} = 0.844\ \text{m}\\ \theta_s &= \frac{L_s}{2R} = \frac{90}{800}\ \text{rad} = 6.446^\circ\\ T_s &= (R+s)\tan 20^\circ + \frac{L_s}{2} = 190.90\ \text{m}\\ L_c &= R(\Delta - 2\theta_s) = 189.25\ \text{m} \end{aligned}$$

Chainages

• Beginning of composite curve, TS $= 1400 - 190.90 = 1209.10$ m
• SC $= 1209.10 + 90 = 1299.10$ m
• CS $= 1299.10 + 189.25 = 1488.36$ m
• End of composite curve, ST $= 1488.36 + 90 = 1578.36$ m

Tangential angles on the transition (from TS)

$\alpha = 573\,l^2/(R L_s)$ minutes. First two points (the first sub-chord runs to the next 30 m peg):

The full transition table is:

Deflection angles on the circular curve (from SC)

$\delta = 1718.87\,c/R$ minutes; first two points:

Full table:

Check: last cumulative angle $= (\Delta - 2\theta_s)/2 = 13.5542^\circ$ = 13°33′15″.

Answer: TS = 1209.10 m and ST = 1578.36 m (beginning and end of the composite curve); deflection angles as tabulated.