Examples & Applications: Complex Vertical Curves

Symmetrical Parabolic Vertical Curves

Example 1: Elevation of Highest Point on a Symmetrical Crest Curve

Problem: A symmetrical parabolic vertical curve has a length L=200 mL = 200\text{ m}. The initial grade is g1=+2%g_1 = +2\% and the final grade is g2=3%g_2 = -3\%. The Point of Vertical Intersection (PVI) is located at station 20+0020+00 with an elevation of 100.00 m100.00\text{ m}. Calculate the station and elevation of the highest point on the curve.
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Example 2: Determining Point of Vertical Tangency (PVT)

Problem: A symmetrical parabolic vertical curve has a length L=150 mL = 150\text{ m}. The initial grade is g1=4%g_1 = -4\% and the final grade is g2=+2%g_2 = +2\%. The Point of Vertical Curvature (PVC) is located at station 10+0010+00 with an elevation of 120.00 m120.00\text{ m}. What is the elevation of the Point of Vertical Tangency (PVT)?
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Example 3: Elevation at Any Point on Symmetrical Curve

Problem: For the same symmetrical vertical curve (Example 2), calculate the vertical offset yy at a horizontal distance x=50 mx = 50\text{ m} from the PVC.
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Unsymmetrical Parabolic Vertical Curves

Example 4: Calculating External Distance for Unsymmetrical Curve

Problem: An unsymmetrical parabolic vertical curve has a total length L=300 mL = 300\text{ m}. The horizontal distance from the PVC to the PVI is L1=120 mL_1 = 120\text{ m} and the distance from the PVI to the PVT is L2=180 mL_2 = 180\text{ m}. The initial grade is g1=2%g_1 = -2\% and the final grade is g2=+3%g_2 = +3\%. What is the external distance EE at the PVI?
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Example 5: Designing an Unsymmetrical Curve to Fit Constraints

Problem: A highway passing under a bridge requires an elevation of at least 105.00 m105.00\text{ m} at the PVI station to maintain clearance. The PVC is at elevation 108.00 m108.00\text{ m} and L1=80 mL_1 = 80\text{ m}. The initial grade is 4%-4\%. What is the maximum length of L2L_2 if the final grade is +2%+2\% to ensure the sag curve clears the bridge constraints? The external distance must not raise the curve above 105.00 m105.00\text{ m} at the PVI.
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Example 6: Evaluating PVI Offset

Problem: For an unsymmetrical curve with L1=150 mL_1 = 150\text{ m}, L2=100 mL_2 = 100\text{ m}, g1=3%g_1 = 3\%, and g2=1%g_2 = -1\%, calculate the difference in elevation between the initial tangent at the PVI and the curve itself at that same station.
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Sight Distance on Vertical Curves

Example 7: Determining Curve Length Based on Stopping Sight Distance (Crest, S < L)

Problem: A crest vertical curve must be designed for a stopping sight distance of S=140 mS = 140\text{ m}. The initial grade is g1=+4%g_1 = +4\% and the final grade is g2=2%g_2 = -2\%. Assuming driver eye height h1=1.08 mh_1 = 1.08\text{ m} and object height h2=0.60 mh_2 = 0.60\text{ m}, determine the required length of the curve. Assume S<LS < L.
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Example 8: Verifying Sight Distance (Sag Curve, S < L)

Problem: A sag vertical curve has a length L=250 mL = 250\text{ m}. The grades are g1=3%g_1 = -3\% and g2=+3%g_2 = +3\%. The stopping sight distance required is S=180 mS = 180\text{ m}. Determine if the curve provides adequate headlight sight distance. Use the standard formula for sag curves L=AS2120+3.5SL = \frac{A S^2}{120 + 3.5S} assuming S<LS < L.
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Example 9: Determining Curve Length Based on Stopping Sight Distance (Crest, S > L)

Problem: A minor crest vertical curve connects an initial grade of g1=+2%g_1 = +2\% and a final grade of g2=1%g_2 = -1\%. The required stopping sight distance is S=200 mS = 200\text{ m}. Assuming driver eye height h1=1.08 mh_1 = 1.08\text{ m} and object height h2=0.60 mh_2 = 0.60\text{ m}, determine the required length of the curve. Assume S>LS > L.
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