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Vertical curve +2% and +5.5%, 60 kmph

Short answer

SSD=82.2 m; headlight criterion with L<S gives L=40 m. BVC elevation =1411.20 m, EVC elevation =1412.70 m.

A vertical curve connects an ascending gradient of +2% grade with another ascending gradient of +5.5% grade on a highway of design speed 60 kmph at an elevation of 1411.6 m. Design the length of the vertical curve and determine the elevations at the beginning and end point of the curve. Use coefficient of longitudinal friction of 0.35.

The grade rises from +2% to the steeper +5.5%, so the curve is a valley (sag) curve. N=g2−g1=0.055−0.02=0.035N=g_2-g_1=0.055-0.02=0.035.

Stopping sight distance (V=60V=60 km/h, f=0.35f=0.35, t=2.5t=2.5 s)

SSD=0.278(60)(2.5)+602254(0.35)=82.2 mSSD=0.278(60)(2.5)+\frac{60^2}{254(0.35)}=82.2\ \text{m}

Length of curve (headlight beam, h1=0.75h_1=0.75 m, α=1∘\alpha=1^\circ)

  • Trial with L>SL>S: L=NS21.5+0.035S=54.0L=\dfrac{NS^2}{1.5+0.035S}=54.0 m, which is less than S=82.2S=82.2 m, so this case does not apply.
  • Case L<SL<S:
L=2S−1.5+0.035SN=2(82.2)−1.5+0.035(82.2)0.035=39.3 mL=2S-\frac{1.5+0.035S}{N}=2(82.2)-\frac{1.5+0.035(82.2)}{0.035}=39.3\ \text{m}
  • Comfort (C=0.6C=0.6 m/s³): L=2NV3/C=32.9L=2\sqrt{NV^3/C}=32.9 m.

Adopt L = 40 m (greater of the two).

Elevations

PVI elevation = 1411.60 m, half length = 20 m.

RLBVC=1411.60−0.02(20)=1411.20 mRLEVC=1411.60+0.055(20)=1412.70 mRL_{BVC}=1411.60-0.02(20)=1411.20\ \text{m}\qquad RL_{EVC}=1411.60+0.055(20)=1412.70\ \text{m}

Answer: L = 40 m; elevation of beginning point = 1411.20 m; elevation of end point = 1412.70 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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