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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

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Introduction to Vertical Curves

CIV2701 Road Design and Location

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

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)

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Vertical Curve Classification

• Usually parabolic as opposed to circular

• Convex (crest curves) or• Concave (sag curves)

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Properties of Parabolic Curve

• Remains a parabola when plotted at exaggerated scale

• Vertical offsets are proportional to square of distance along tangent

• A point on parabola lies halfway along the line from IP to mid point on chord

• Vertical acceleration is constant• For flat gradient curves it is

assumed that length of chord=arc length=sum of tangent lengths = distance between tangent points

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Note: L is measured from here to here

Not here

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Basic Formulae

Equation for Parabola y = kx2

Slope at any point dy/dx = 2kxRate of change of slope = d2y/dx2 = 2k

g1 = grade 1g2 = grade 2A = difference in grade = g2 – g1

L = length of curveK = L/A = rate of vertical curvature

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Computations on the Vertical Curve

Key Formulae

Equation for Parabola y = kx2

Equivalent Radius =R = 100 L/AVertical offset = y =Ax2/200LMid-ordinate = e = LA/800RL at any point = RLTP + xg1/100 – y

Distance to highest (or lowest point) = x = Lg1/A

This distance is from TP1

A similar calculation can be done from TP2 where x= Lg2/A

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Vertical Curve Components

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

ExampleA crest vertical curve joins a +3% and –4% grade. Design speed is 100km/hr. Length = 530m. The chainage at the TP is 3460.00m, RL of 52.50m

Calculate points along the vertical curve at chainage3500.0, 3600 and 3700m

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

For Chainage 3500m

X = 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= 40.0m Y= 7%*402/200*530 =0.106So RL @ 3500m = 52.50+ 40*3/100 -0.106

= 53.594m

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

For Chainage 3600m

X = 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= 140.0m Y= 7%*1402/200*530 = 1.294So RL @ 3600m = 52.50+ 140*3/100 – 1.294

= 55.406m

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

For Chainage 3700m

X = 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= 240.0m Y= 7%*2402/200*530 = 3.804So RL @ 3700m = 52.50+ 240*3/100 – 3.804

= 55.896m

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

ExampleA crest vertical curve joins a +3% and –4% grade. Design speed is 100km/hr. Length = 530m. The chainage at the TP is 3460.00m, RL of 52.50m

Compute the RL and chainage of the highest point on the curve.

T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Review Formulae

Key Formulae

Equation for Parabola y = kx2

Equivalent Radius =R = 100 L/AVertical offset = y =Ax2/200LMid-ordinate = e = LA/800RL at any point = RLTP + xg1/100 – y

Distance to highest (or lowest point) = x = Lg1/AThis distance is from TP1

A similar calculation can be done from TP2 where x= Lg2/A

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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d

Compute Highest Point

Distance to highest (or lowest point) = x = Lg1/AThis distance is from TP1

So, 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