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Vertical curve elevations at 185 m from PVI

Short answer

Summit curve with grades +3% and -2.5%, radius 8500 m, PVI elevation 500 m. Curve length 467.5 m. Using parabolic formula, elevation 185 m before PVI is 494.31 m and 185 m after PVI is 495.24 m.

A ascending gradient 3% meets the 2.5% descending gradient. Calculate the elevation of the road at a distance of every 185 m on either side of the point of vertical intersection (PVI) if the elevation of PVI is 500.00 m. Assume the radius of vertical curve as 8500 m.

A +3% grade followed by a −2.5% grade is a summit curve. N=0.03+0.025=0.055N=0.03+0.025=0.055, R=8500R=8500 m, PVI RL = 500.00 m.

Length of curve

L=NR=0.055(8500)=467.5 m,L2=233.75 mL=NR=0.055(8500)=467.5\ \text{m},\qquad \frac{L}{2}=233.75\ \text{m}

Since 185 m < 233.75 m, both required points lie on the curve.

Levels of BVC and EVC

RLBVC=500.00−0.03(233.75)=492.9875 m,RLEVC=500.00−0.025(233.75)=494.1563 mRL_{BVC}=500.00-0.03(233.75)=492.9875\ \text{m},\qquad RL_{EVC}=500.00-0.025(233.75)=494.1563\ \text{m}

Elevation 185 m before PVI (on the BVC side)

Distance from BVC: x=233.75−185=48.75x=233.75-185=48.75 m.

y=RLBVC+g1x−x22R=492.9875+0.03(48.75)−48.7522(8500)=494.310 my=RL_{BVC}+g_1x-\frac{x^2}{2R}=492.9875+0.03(48.75)-\frac{48.75^2}{2(8500)}=494.310\ \text{m}

Elevation 185 m after PVI (on the EVC side)

Distance from EVC: x′=48.75x'=48.75 m (measured back from EVC along the falling 2.5% grade):

y=RLEVC+g2x′−x′22R=494.1563+0.025(48.75)−48.7522(8500)=495.235 my=RL_{EVC}+g_2x'-\frac{x'^2}{2R}=494.1563+0.025(48.75)-\frac{48.75^2}{2(8500)}=495.235\ \text{m}

(Check: tangent level 185 m before PVI is 500−0.03(185)=494.450500-0.03(185)=494.450 m; the curve is below it by x2/2R=0.140x^2/2R=0.140 m, giving 494.310 m.)

Answer: elevation 185 m before PVI = 494.31 m; elevation 185 m after PVI = 495.24 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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