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Vertical Curves

Vertical Curves. Sometimes required when two gradients meet Transition too abrupt Sight Distances Note terms Distances measured horizontally Tangent offsets indicate cut or fill from flat grade to curve. Parabolic Curve. Form: y = ax 2 + bx +c Three important relationships

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Vertical Curves

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  1. Vertical Curves • Sometimes required when two gradients meet • Transition too abrupt • Sight Distances • Note terms • Distances measured horizontally • Tangent offsets indicate cut or fill from flat grade to curve

  2. Parabolic Curve • Form: y = ax2 + bx +c • Three important relationships • Curve at midpoint is always halfway between PVI and a line connecting PVC and PVT. • Tangent offsets vary with the square of the distance from PVC. • Second differences between equal points are equal.

  3. Parabolic Example • Extend g1, divide into 6 equal parts • If offset at 1 = “a” • Offset2 = 4a • Offset3 = 9a • ACV similar to ABB’ • CV = BB’/2 = 18a • CM = MV = 9a

  4. Vertical Curve by Proportions • Problem 21-3 • g1 = 4%, g2 = -2.1% • PVT = 46+50, E = 682.24 • L = 8 sta (800 ft) • Sta PVC = Sta PVI – L/2 = 42+50, EPVC = 682.24 – 4%(4 Sta) = 666.24 • Sta PVT Sta PVI + L/2 = 50+50, EPVT = 682.24 + 4 Sta(-2.1%) = 673.84

  5. Problem 21-3 (Cont.) • Find yPVT, yMid • yPVT = EPVC + g1L – EPVT = 666.24 +(4%)(8 Sta) – 673.84 = 24.40’ • yMid = yPVT(4/8)2 = 24.40’(.5)2 = 6.10’, or

  6. Problem 21-3, Cont. • Ascending: E = EPVC + g1x + y • y43 = (-6.10)(0.5)2/(4)2 = -0.10 • E43 = 666.24 +(4%)(0.5 sta) – 0.10 = 668.14 • y44 = (-6.10)(1.5)2/(4)2 = -0.86 • E43 = 666.24 +(4%)(1.5 sta) – 0.86 = 671.38 • Descending: E = EPVT – g2x + y • Sta 47+00, x from PVT = 3.5 sta • y47 = (-6.10)(3.5)2/(4)2 = -4.67 • E47 = 673.84 - (-2.1%)(3.5 sta) – 4.67 = 676.52

  7. Curve Equation • Elev = c + bx +ax2 • c = ElevA • b = g1 (in %) • x in Sta • ElevX = ElevA+g1x+ax2 • Rate of Change • d2y/dx2 = 2a • a = (g2-g1)/2L

  8. Vertical Curve Equation • Change in grade: A = g2-g1 • Rate of change: r = A/L • High or Low Point • Occurs when grade = 0 • Grade changes from g1 to 0 at rate r

  9. Problem 21-3 by Equation • A = (-2.1% – 4%) = -6.1% • r = A/L = -6.1%/8 Sta = -.7625 %/Sta • E43 = 666.24 + 4(0.5) - 0.38125(.5)2 = 668.14 • E44 = 666.24 + 4(1.5) - 0.38125(1.5)2 = 671.38 • E46+50 = 666.24 + 4(4.0) - 0.38125(4.0)2 = 676.14 • E47 = 666.24 + 4(4.5) - 0.38125(4.5)2 = 676.52 • High Point: X = -g1/r = -4/-0.7625 = 5.2469 Sta • E47+74.59 = 666.24 + 4(5.2469) - .38125(5.2469)2 = 676.73 • E48 = 666.24 + 4(5.5) - 0.38125(5.5)2 = 676.71

  10. Curve Thru a Point • Say you want the curve to pass through a point of known Sta, Elev • Select L to force curve through point • Two possible solutions • Tangent Offsets – Author • Equation

  11. Thru Point - Tangent Offset • xP = Distance from PVI to Point, in Sta • yB = offset from the back tangent • yF = offset from forward tangent

  12. Thru Point – Tangent Offset

  13. Thru Point – Equation • yB – yF = (g2 - g1)xP • yB + yF = 2(EP - EPVI) - (g2 + g1)xP • Recall A = g2 - g1 • xP = Distance from PVI to point, in Sta

  14. Superelevation • Crown: 0.015 – 0.02 ft/ft • Superelevation • Transverse Slope • Depends on Side friction, speed, D • Max: 8%- 10% • Runoff – transition from flat to banked • Linear change • Min: 100-250 ft • Max: 500-600 ft

  15. Superelevation Example • Road conditions • Curve, D = 4° , I = 53° • PC at Sta 24+45, Elev = 299.75’ • Centerline grade from PC to PT is –3.5% • Road top is 40’ wide, crown is 0.015 ft/ft • Superelevation = 4%, Runoff = 400 feet • Design • PC Centerline at 299.75’ • At PC need full superelevation,  .04(20’) = 0.80’ • Out shoulder at 300.55’, In shoulder at 298.95’

  16. Runoff • PC at Sta 24+45, Elev = 299.75’ • Out shoulder at 300.55’, In shoulder at 298.95’ • 400 foot runoff – start at Sta 20+45 • Centerline at 313.75 • Crown at 0.015*20’ = 0.30’ • Out and In Shoulders at 313.45 • Linear transition • Roll outside crown up to 1.5% (0.015 ft/ft) • Roll both shoulders to 4%

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