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Transition curve 6°, Ls=90 m, chainage (13+7) chains

Short answer

Deflection angle Δ = 40° right. R = 1718.87/6 = 286.48 m; transition length Ls = 90 m; shift s = 1.178 m; tangent length Ts = 149.70 m; circular curve length Lc = 110.00 m. Chainages: TS 247.30 m, SC 337.30 m, CS 447.30 m, ST 537.30 m.

Two tangents T₁V and T₂V intersect at chainage of (13+7) Chains. The bearing of forward tangent T₂V is 300°30' and backward tangent T₁V is 80°30'. It is proposed to insert 6° Circular Curve with transition curve 90 m at each end. Make all the calculations necessary for setting out the curve at peg interval of 15 m on the transition curve and 30 m at the circular curve.

Deflection angle between the tangents

Bearing of T₁V (travelling towards V) =80∘30′= 80^\circ30'. Bearing of T₂V is observed from T₂ towards V, 300∘30′300^\circ30', so the forward direction of travel from V to T₂ is 300∘30′−180∘=120∘30′300^\circ30' - 180^\circ = 120^\circ30'.

Δ=120∘30′−80∘30′=40∘ (deflection to the right)\Delta = 120^\circ30' - 80^\circ30' = 40^\circ\ \text{(deflection to the right)}

Radius and transition elements

6° curve with a 30 m arc: R=1718.87/6=286.48R = 1718.87/6 = 286.48 m. Ls=90L_s = 90 m.

s=Ls224R=90224×286.48=1.178 mθs=Ls2R=9.000∘Ts=(R+s)tan⁡20∘+Ls2=149.70 mLc=R(Δ−2θs)=110.00 m\begin{aligned} s &= \frac{L_s^2}{24R} = \frac{90^2}{24\times 286.48} = 1.178\ \text{m}\\ \theta_s &= \frac{L_s}{2R} = 9.000^\circ\\ T_s &= (R+s)\tan 20^\circ + \frac{L_s}{2} = 149.70\ \text{m}\\ L_c &= R(\Delta-2\theta_s) = 110.00\ \text{m} \end{aligned}

Chainage of V =397= 397 m ((13+7) chains read as 13 × 30 + 7 m).

PointChainage (m)
TS = V − T_s247.30
SC = TS + 90337.30
CS = SC + L_c447.30
ST = CS + 90537.30

Transition curve (theodolite at TS), peg interval 15 m

α=573 l2/(RLs)\alpha = 573\,l^2/(RL_s) minutes:

Point (chainage, m)l from TS (m)α (min)α (° ′ ″)
247.300.000.00′0°00′00″
255.007.701.32′0°01′19″
270.0022.7011.45′0°11′27″
285.0037.7031.58′0°31′35″
300.0052.7061.72′1°01′43″
315.0067.70101.85′1°41′51″
330.0082.70151.99′2°31′59″
337.3090.00180.01′3°00′01″

Circular curve (theodolite at SC), peg interval 30 m

δ=1718.87 c/R\delta = 1718.87\,c/R minutes:

Point (chainage, m)Chord c (m)δ = 1718.87c/RCumulative deflection
337.30--0°00′00″
360.0022.702°16′11″2°16′11″
390.0030.003°00′00″5°16′11″
420.0030.003°00′00″8°16′11″
447.3027.302°43′48″11°00′00″

Check: last cumulative deflection =(Δ−2θs)/2=11.0000∘= (\Delta - 2\theta_s)/2 = 11.0000^\circ = 11°00′00″; at SC the transition angle αs=θs/3=3.0000∘\alpha_s = \theta_s/3 = 3.0000^\circ = 3°00′00″. The second transition is set out from ST backwards, in the same way.

Answer: Δ=40∘\Delta = 40^\circ, R=286.48R = 286.48 m, Ts=149.70T_s = 149.70 m, Lc=110.00L_c = 110.00 m; TS = 247.30, SC = 337.30, CS = 447.30, ST = 537.30 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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