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

3. 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 )

4. Vertical Curve Classification Usually parabolic as opposed to circular Convex (crest curves) or Concave (sag curves)

6. Properties of Parabolic Curve Remains a parabola when plotted at exaggerated scale Vertical offsets are proportional to square of distance along tangent

7. Vertical acceleration is constant For flat gradient curves it is assumed that length of chord=arc length=sum of tangent A point on parabola lies halfway along the line from IP to mid point lengths = distance between tangent points

8. Basic Formulae Equation for Parabola y = kx2 Slope at any point dy/dx = 2kx Rate of change of slope = d2y/dx2 = 2k g1 = grade 1 g2 = grade 2 A = difference in grade = g2 – g1 L = length of curve K = L/A = rate of vertical curvature

9. Computations on the Vertical Curve Key Formulae Equation for Parabola y = kx2 Equivalent Radius =R = 100 L/A Vertical offset = y =Ax2/200L Mid-ordinate = e = LA/800 RL 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

10. Example A 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 chainage 3500.0, 3600 and 3700 m

11. For Chainage 3500m X = distance from TP Y = Ax2/200 L RL at any point = RLTP + xg1/100 – y A=g2-g2 = -4-3 = -7% = 7% (ignore sign) So for chainage 3500 X= 40.0m Y= 7%*402/200*530 =0.106 So RL @ 3500m = 52.50+ 40*3/100 -0.106 = 53.594m

12. For Chainage 3600m X = distance from TP Y = Ax2/200 L RL at any point = RLTP + xg1/100 – y A=g2-g2 = -4-3 = -7% = 7% (ignore sign) So for chainage 3600 X= 140.0m Y= 7%*1402/200*530 = 1.294 So RL @ 3600m = 52.50+ 140*3/100 – 1.294 = 55.406m

13. For Chainage 3700m X = distance from TP Y = Ax2/200 L RL at any point = RLTP + xg1/100 – y A=g2-g2 = -4-3 = -7% = 7% (ignore sign) So for chainage 3700 X= 240.0m Y= 7%*2402/200*530 = 3.804 So RL @ 3700m = 52.50+ 240*3/100 – 3.804 = 55.896m

14. Compute Highest Point Distance to highest (or lowest point) = x = Lg1/A This distance is from TP1 So, X= 530*3/7 =227.143 Chainage of point = TP1 + x = 3460 + 227.143 = 3687.143m Then Y = 7%*227.1432/200*530 = 3.408 So RL @ 3687.143m = 52.50+ 227.143*3/100 – 3.408 = 55.907m