Curve by tangential angle, deflection 48°30'

Data: $\Delta = 48^\circ30'$, chainage of IP = 1+945.55 (1945.55 m), tangent length not more than 35 m, normal chord $C = 20$ m.

1. Selecting the radius

$$T = R\tan\frac{\Delta}{2} \le 35 \quad\Rightarrow\quad R \le \frac{35}{\tan 24^\circ15'} = 77.70\ \text{m}$$

Adopt R = 75 m (the next lower round value). Then

$$T = 75\tan 24^\circ15' = 33.785\ \text{m} \ (< 35\ \text{m})$$

2. Curve elements

$$L_c = \frac{\pi R\Delta}{180^\circ} = \frac{\pi\times 75\times 48.5}{180} = 63.486\ \text{m}$$

• Chainage of T1 $= 1945.55 - 33.785 = 1911.765$ m
• Chainage of T2 $= 1911.765 + 63.486 = 1975.251$ m

3. Setting out table (tangential angles)

$$\delta = \frac{1718.9\,c}{R}\ \text{minutes} \quad (\delta_{20} = 7°38'22")$$

Pegs are at the multiples of 20 m; the first and the last chords are sub-chords.

Check: the last cumulative angle 24°15'01" $= \Delta/2 = 24^\circ15'00''$ ✓.

Procedure: set up at T1, set the horizontal circle to zero on the IP, and set out each peg by turning the cumulative angle in the table and measuring the chord from the previous peg with a tape; the last peg must fall on T2, which checks the work.

Answer: R = 75 m, T = 33.79 m, $L_c$ = 63.49 m, T1 = 1911.76 m, T2 = 1975.25 m.