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Simple circular curve elements and chainages

Short answer

T = 83.87 m, Lc = 158.82 m, C = 154.68 m, E = 16.87 m, M = 15.56 m. Chainage of PC = 20832.73 m, PT = 20991.56 m.

At a certain section of road there is an intersection point (IP) with an angle of 45°30' turning right. The minimum permissible radius is 200 m. The distance between starting point and IP and IP to end point are 500 and 350 m respectively. Calculate the elements of circular curve and chainages of main points of curve assuming the chainage of starting point is to be 20 + 416.60.

Given

Deflection angle Δ=45∘30′=45.5∘\Delta=45^\circ30'=45.5^\circ (right turn), R=200R=200 m. Start chainage = 20+416.60 = 20 416.60 m; start to IP = 500 m; IP to end = 350 m.

Elements of the simple circular curve

  • Tangent length: T=Rtan⁡Δ2=200tan⁡22.75∘=83.87T=R\tan\dfrac{\Delta}{2}=200\tan22.75^\circ=\mathbf{83.87} m
  • Length of curve: Lc=πRΔ180=π(200)(45.5)180=158.82L_c=\dfrac{\pi R\Delta}{180}=\dfrac{\pi(200)(45.5)}{180}=\mathbf{158.82} m
  • Long chord: C=2Rsin⁡Δ2=2(200)sin⁡22.75∘=154.68C=2R\sin\dfrac{\Delta}{2}=2(200)\sin22.75^\circ=\mathbf{154.68} m
  • External distance: E=R(sec⁡Δ2−1)=16.872E=R\left(\sec\dfrac{\Delta}{2}-1\right)=\mathbf{16.872} m
  • Mid-ordinate: M=R(1−cos⁡Δ2)=15.560M=R\left(1-\cos\dfrac{\Delta}{2}\right)=\mathbf{15.560} m

Chainages

  • Chainage of IP =20 416.60+500=20916.60=20\,416.60+500=20916.60 m (20+916.60)
  • Beginning of curve (PC) =IP−T=20916.60−83.87=20832.73=IP-T=20916.60-83.87=\mathbf{20832.73} m
  • Mid-point of curve =PC+Lc/2=20912.15=PC+L_c/2=\mathbf{20912.15} m
  • End of curve (PT) =PC+Lc=20832.73+158.82=20991.56=PC+L_c=20832.73+158.82=\mathbf{20991.56} m
  • End of the road =PT+(350−T)=20991.56+266.13=21257.69=PT+(350-T)=20991.56+266.13=\mathbf{21257.69} m

The minimum permissible radius is 200 m, and the curve uses exactly this radius.

✓ Answer checked by Claude Opus 5.5 (high effort)

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