a curve inserted between two lengths of a road or railway which are at different slopes. vertical...

A curve inserted between two lengths of a road or railway which are at different slopes.

Vertical curve

vertical curveA smooth parabolic curve in the vertical plane used to connect two grades of different slope to avoid an abrupt transition in passing from one to the other.

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Purpose of Vertical Curves • Allow smooth transition from one grade to another (driver comfort) • Provide adequate sight distance at junction of grades and for overtaking (safety) • Provide satisfactory appearance (aesthetics)

Vertical Curve Classification• Usually parabolic asopposed to circular• Convex (crest curves) or• Concave (sag curves)

Properties of Parabolic Curve• Remains a parabola when plottedat exaggerated scale• Vertical offsets are proportional tosquare of distance along tangent

• Vertical acceleration is constant• For flat gradient curves it isassumed that length of chord=arclength=sum of tangent

• A point on parabola lies halfwayalong the line from IP to mid point

lengths =distance between tangent points

Basic FormulaeEquation for Parabola y = kx2Slope at any point dy/dx = 2kxRate of change of slope = d2y/dx2 = 2kg1 = grade 1g2 = grade 2A = difference in grade = g2 – g1L = length of curveK = L/A = rate of vertical curvature

Computations on the Vertical CurveKey FormulaeEquation for Parabola y = kx2Equivalent Radius =R = 100 L/AVertical offset = y =Ax2/200LMid-ordinate = e = LA/800RL at any point = RLTP + xg1/100 – yDistance to highest (or lowest point) = x =Lg1/AThis distance is from TP1A similar calculation can be done from TP2where x= Lg2/A

ExampleA crest vertical curve joins a +3% and –4% grade.Design speed is 100km/hr. Length = 530m. Thechainage at the TP is 3460.00m, RL of 52.50mCalculate points along the vertical curve at chainage3500.0, 3600 and 3700m

For Chainage 3500mX = distance from TPY = Ax2/200 LRL at any point = RLTP + xg1/100 – yA=g2-g2 = -4-3 = -7% = 7% (ignore sign)So for chainage 3500X= 40.0mY= 7%*402/200*530 =0.106So RL @ 3500m = 52.50+ 40*3/100 -0.106= 53.594m

For Chainage 3600mX = distance from TPY = Ax2/200 LRL at any point = RLTP + xg1/100 – yA=g2-g2 = -4-3 = -7% = 7% (ignore sign)So for chainage 3600X= 140.0mY= 7%*1402/200*530 = 1.294So RL @ 3600m = 52.50+ 140*3/100 – 1.294= 55.406m

For Chainage 3700mX = distance from TPY = Ax2/200 LRL at any point = RLTP + xg1/100 – yA=g2-g2 = -4-3 = -7% = 7% (ignore sign)So for chainage 3700X= 240.0mY= 7%*2402/200*530 = 3.804So RL @ 3700m = 52.50+ 240*3/100 – 3.804= 55.896m

Compute Highest PointDistance to highest (or lowest point) = x = Lg1/AThis distance is from TP1So, X= 530*3/7 =227.143Chainage of point = TP1 + x = 3460 + 227.143= 3687.143mThen Y = 7%*227.1432/200*530 = 3.408So RL @ 3687.143m = 52.50+ 227.143*3/100 – 3.408= 55.907m