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Summit curve +1.5% and -2%

Short answer

For N = 0.035, SSD 200 m gives L = NS^2/4.4 = 318 m, vertical offset at PVI E = NL/8 = 1.39 m. OSD 600 m gives L = NS^2/9.6 = 1313 m, E = 5.74 m.

A summit curve is to be provided at the intersection of two gradients +1.5% and -2%. What is length required (i) For stopping sight distance of 200 m (ii) For overtaking sight distance of 600 m? What is the vertical distance between the point of vertical intersection and curve in either case?

g1=+1.5%, g2=−2%g_1=+1.5\%,\ g_2=-2\%, so N=0.015+0.02=0.035N=0.015+0.02=0.035 (summit curve). Eye height 1.2 m, object height 0.15 m for SSD and 1.2 m for OSD (IRC/NRS 2070).

(i) SSD = 200 m

Assuming L>SL>S:

L=NS24.4=0.035(200)24.4=318.2 mL=\frac{NS^2}{4.4}=\frac{0.035(200)^2}{4.4}=318.2\ \text{m}

Since L>SL>S, it is valid. L = 318 m (say 320 m).

(ii) OSD = 600 m

Here h1=h2=1.2h_1=h_2=1.2 m:

L=NS29.6=0.035(600)29.6=1312.5 mL=\frac{NS^2}{9.6}=\frac{0.035(600)^2}{9.6}=1312.5\ \text{m}

L = 1313 m (>S>S, valid).

Vertical distance between PVI and curve

For a parabolic curve the offset at the PVI is E=NL8E=\dfrac{NL}{8}.

  • (i) E=0.035(318.2)8=1.392E=\dfrac{0.035(318.2)}{8}=\mathbf{1.392} m
  • (ii) E=0.035(1312.5)8=5.742E=\dfrac{0.035(1312.5)}{8}=\mathbf{5.742} m

Answer: (i) L = 318 m, E = 1.39 m; (ii) L = 1313 m, E = 5.74 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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