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Combined curve table, R=300 m, Ls=60 m

Short answer

Compute shift s, spiral angle θs, tangent length Ts, circular length Lc. Ts = 154.47 m, Lc = 175.62 m. Chainages: TS = 1845.53 m, SC = 1905.53 m, CS = 2081.15 m, ST = 2141.15 m. Set out transitions with α = 573l²/(RLs) min and circular curve with δ = 1718.87c/R min.

It is proposed to insert a circular curve of 300 m radius with a transition curve of length 60 m long each end of the circular curve. Prepare necessary data for setting out the combined curve in tabular form. Deflection angle between two alignments of road is 45° and chainage of intersection point is 2000 m. Peg interval for transition and circular curve are 20 m and 30 m respectively. Take chainage at multiple of peg interval.

Given Δ=45∘\Delta = 45^\circ, R=300R = 300 m, Ls=60L_s = 60 m, chainage of IP = 2000 m, peg interval 20 m on transitions and 30 m on the circular curve.

Curve elements

s=Ls224R=60224×300=0.500 mθs=Ls2R rad=5.730∘Ts=(R+s)tan⁡22.5∘+Ls2=154.47 mLc=πR180(Δ−2θs)=175.62 m\begin{aligned} s &= \frac{L_s^2}{24R} = \frac{60^2}{24\times 300} = 0.500\ \text{m}\\ \theta_s &= \frac{L_s}{2R}\ \text{rad} = 5.730^\circ\\ T_s &= (R+s)\tan 22.5^\circ + \frac{L_s}{2} = 154.47\ \text{m}\\ L_c &= \frac{\pi R}{180}(\Delta-2\theta_s) = 175.62\ \text{m} \end{aligned}
PointChainage (m)
TS = 2000 − 154.471845.53
SC = TS + 601905.53
CS = SC + 175.622081.15
ST = CS + 602141.15

First transition (set from TS): α=573 l2/(RLs)\alpha = 573\,l^2/(RL_s) minutes

Point (chainage, m)l from TS (m)α (min)α (° ′ ″)
1845.530.000.00′0°00′00″
1860.0014.476.67′0°06′40″
1880.0034.4737.83′0°37′50″
1900.0054.4794.45′1°34′27″
1905.5360.00114.60′1°54′36″

Check: at SC, α=θs/3=1.9099∘=1∘54′36′′\alpha = \theta_s/3 = 1.9099^\circ = 1^\circ54'36''.

Circular curve (set from SC): δ=1718.87 c/R\delta = 1718.87\,c/R minutes

Point (chainage, m)Chord c (m)δ = 1718.87c/RCumulative deflection
1905.53--0°00′00″
1920.0014.471°22′55″1°22′55″
1950.0030.002°51′53″4°14′48″
1980.0030.002°51′53″7°06′41″
2010.0030.002°51′53″9°58′34″
2040.0030.002°51′53″12°50′28″
2070.0030.002°51′53″15°42′21″
2081.1511.151°03′52″16°46′13″

Check: last cumulative deflection =(Δ−2θs)/2=16.7704∘= (\Delta - 2\theta_s)/2 = 16.7704^\circ = 16°46′13″. The second transition is set out from ST in the same way, using ll measured back from ST.

Answer: Ts=154.47T_s = 154.47 m, Lc=175.62L_c = 175.62 m, TS = 1845.53, SC = 1905.53, CS = 2081.15, ST = 2141.15 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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