Lecture 3 - Vertical Curves 1w5c2i

Surveying-II CE-207 (T) Curves Vertical curves Lecture 3

Engr. Awais Khan

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Curves Vertical Curves • In addition to horizontal curves that go to the right or left, roads also have vertical curves that go up or down. • Vertical curves at a crest or the top of a hill are called summit curves, or oververticals. • Vertical curves at the bottom of a hill or dip are called sag curves, or underverticals.

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Curves Vertical Curves • In the same way that horizontal curves are used to connect intersecting straights in the horizontal plane, vertical curves are used to connect intersecting straights in the vertical plane. • These straights are usually called gradients and the combination of the of the gradients and vertical curve is known as vertical alignment.

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Curves Vertical Curves • The vertical alignment is composed of a series of straightline gradients connected by curves, normally parabolic in form. • These vertical parabolic curves must therefore be provided at all changes in gradient. • The curvature will be determined by the design speed, being sufficient to provide adequate driver comfort with appropriate stopping sight distances provided.

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Curves Vertical Curves Purpose of Vertical Curves • Vertical curves are similar to horizontal curves in that they are designed for a particular speed. The main functions of a vertical curve are to provide: • Safety and comfort travelling between gradients • Adequate visibility to enable vehicles to stop or overtake safely.

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Curves Vertical Curves • When two different gradients meet they are connected by a curve in a vertical plane is called vertical curve. it is used in roads and railway. Objective: • To round off the angle at the apex and to obtain a gradual change in grade or gradient, so that abrupt change in grade is avoided at the apex. • Vertical curves may be: a) Circular arc

b) Arc of Parabola

• For simplicity of calculation work, the parabola is preferred. •

When the ratio of length of curve to radius is less than 1 to 10, i.e 𝐿 < 1 to 10 𝑅

•

Than there is no practical difference b/w shapes of parabola, circle and an ellipse, the parabola will be used.

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Curves Vertical Curves • In practice, road and railway gradients are comparatively flat and it is often unimportant what type of vertical curve is used. • The usual curves are circular or parabolic. However, it is best to use a vertical curve having a constant rate of change of gradient, i.e., a parabola and as it turns out, parabolic vertical curves are very easy to calculate and use.

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Curves Vertical Curves • Parabola has desirable characteristics of: 1) Constant rate of change of slopes which contributes to sooth alignment transition. 2) Ease of computation of vertical offsets which permits easily computed curve. • The Grade for the gradient of a road is expressed in two ways:

a) As a percentage i.e 2 % 1.5 % b) As 1 in n , where n is horizontal distance in meter or feet corresponding to 1m or 1ft rise or fall. E.g 1 in 80 & 1 in 100.

1 in 50 = 2%, 1 in 25 = 4%, Sign convention: a) Gradient rising ( ascending) to the right are taken +ve b) Gradient falling ( descending) to right are taken –ve

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Curves Vertical Curves Types of vertical curve: 1) An up grade followed

by a down grade

Summit or Convex

2) A down grade followed

by an Up grade

Sag or Concave

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Curves Vertical Curves Types of vertical curve: 3) An up grade followed

4) A down grade followed

by another up grade

by another down grade

Sag or Concave

Summit or Convex

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Curves Vertical Curves Types of parabola curves: Two basic types

1) Cest

2) Sag

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Curves Vertical Curves Assumptions in Vertical Curve Calculations • • • • • •

Choosing a parabola simplifies the calculations of a vertical curve, further simplifications are also made if certain assumption are used. Vertical curves are generally designed with equal tangent lengths such that PQ = OR. Assumptions made are: Chord PWR = Arc PSR = PQ + QR Length along the tangents = horizontal length, i.e. PQ = PQ’ QU = QW, i.e. there is no difference in dimensions measured either in the vertical plane or perpendicular to the entry tangent length

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Curves Vertical Curves Properties of Parabola a) The vertical through the intersection of tangent A, bisect OB. b) OA = AB and FA = FE, where F is vertex of parabola c) Offset y from tangent OA are proportional to the square of distance x from O. The offsets should be perpendicular to the tangent but as flat gradients are usually involved it is used sufficiently accurate to take them vertically. 13

Curves Vertical Curves Properties of Parabola Let eqn of parabola is 𝑦 = 𝐾 𝑥2 𝑑𝑦 𝑑𝑥

=2Kx

𝑑𝑦 2 𝑑𝑥 2

=2K

(gradient slope) (a constant)

i.e rate of change of slope is constant. Thus parabola gives an even rate of gradient. For flat gradients it is usually accurate enough to treat length along tangent be equal to the horizontal projection of tangent. 14

Curves Vertical Curves Properties of Parabola a) To find K, at x = L 𝑔

𝐿

𝑦 = 𝐵𝐶 = 1 x + (− 100 2 𝑔1 − 𝑔2 𝐿 𝑦= 200

𝑔2 𝐿 x ) 100 2

From eqn y = k x2 𝑔1 −𝑔2 𝐿 200 (𝑔1

𝐾= 𝑦=

= 𝐾 𝑥 L2 − 𝑔2 ) 200𝐿

𝑔1 −𝑔2 200𝐿

𝑥2

General form of Parabolic equation 15

Curves Vertical Curves Properties of Parabola 𝐿

b) At x = 2 , y= AF (from tangent to curve) 𝑦=

𝑔1 −𝑔2 200𝐿

𝐿

𝑦=

𝑔1 − 𝑔2 𝐿 800

(2)2

From similar ∆s OAE ,OCB 𝐸𝐴 𝐵𝐶 = 𝐿/2 𝐿 𝐸𝐴 =

𝐵𝐶 2

,

y = BC 1 𝑔1 − 𝑔2 𝐿 𝑔1 − 𝑔2 𝐿 𝐸𝐴 = 𝑥 = = 2 𝐴𝐹 2 (200) (400) i.e AF = FE=

𝑔1 − 𝑔2 800

L

In vertical curve we have horizontal distance and offsets only. 16

Curves Vertical Curves Location of Highest and Lowest Point Taking O as datum any point on the curve will have height 𝐻=𝑦=

𝑔1 𝑥 100

−

𝑔1 − 𝑔2 𝑥2 200𝐿

For max value of H 𝑑𝐻 =0 𝑑𝑥 𝑑𝐻 𝑔1 𝑔1 − 𝑔2 = − 𝑥 𝑑𝑥 100 100𝐿

𝑥=

𝑔1 𝐿 , (𝑔1 − 𝑔2 )

=0

gives distance at point where H is

maximum , applicable for both crest as well as sag. (𝑔1 % − 𝑔2 %) Length of curve = 𝐿 = 𝑟% r = rate of change of gradients = 0.1 per 100 ft or 30m. If r is given then 𝑥 =

𝑔1 𝐿 𝑟

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Curves Vertical Curves Length of Vertical Curve Length of vertical curve is influenced by

1) Centrifugal effect 2) Visibility At sags and at summit, formed by flat gradients, centrifugal effect is the chief consideration.

while at summit where the algebraic change of gradient is large, visibility is main concern. 1) Centrifugal effect:

A minimum radius of 1000 m should be used at sags and brows. This gives a centrifugal acceleration of 0.75 𝑚 𝑠𝑒𝑐2 at 100 𝑘𝑚 ℎ . 18

Curves Vertical Curves Length of Vertical Curve 1) Centrifugal effect: Parabola can be approximate to circle, 𝑑𝑦 2 𝑑𝑥 2

1 𝑅

=2K = ,

𝐾=

(𝑔1 −𝑔2 ) 200𝐿

1 2 (𝑔1 − 𝑔2 ) = 𝑅 200 𝐿 𝑅=

100 𝐿 (𝑔1 −𝑔2 )

,

L=

𝐿𝑚𝑖𝑛 (at R = 1000 m) =

𝑅 (𝑔1 −𝑔2 ) 100 1000 𝑔1 −𝑔2 100

, 𝐿𝑚𝑖𝑛 = 10 (𝑔1 − 𝑔2 )

On highways (𝑔1 −𝑔2 ) 𝑆 2 800 ℎ

,𝐿= Where S = sight distance, h =height of drivers eye above road , h = 1.15m ( NAASRA), h = 1.50 (AASHTO), h= 1.1 m (book)

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Curves Vertical Curves Length of Vertical Curve 2) Sight distance:

At summits where speed of 100 km/ hr are contemplated, the requirement of visibility 1.e the sight line will lead to longer curves than one obtained by above formula. In the design of summit or convex curve it is required to calculate the minimum length of curve which will give the required sight distance. “ The length of road visible to the driver clear of object while driving is called sight distance”.

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Curves Vertical Curves Length of Vertical Curve 2) Sight distance:

Three cases will be considered: 1) The length of curve is grater than the sight distance ( L > S) 2) The length of curve is smaller than the sight distance ( L < S) 3) The length of curve is equal to required sight distance ( L = S) Assignment

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Curves Vertical Curves Procedure for Computing a Vertical Curve Two methods

1) Tangent corrections or tangent offset Method 2) By Chord Gradients 1) Tangent corrections or tangent offset Method

a) Determined gradients and compute the curve length L b) Compute the chainage of tangent points O and B, if chainage of intersection point A is given 𝐿

Chainage of O = chainage of A - 2 𝐿

Chainage of B = chainage of A + 2 22

Curves Vertical Curves Procedure for Computing a Vertical Curve 1) Tangent corrections or tangent offset Method c) Compute the elevation of tangent point O and B from the elevation of intersection point A Elevation of O = elevation of A +

g1 L

Elevation of B = elevation of A -

g2 L

200

200

d) Compute the tangent offset for the stations or pegs on the curve Tangent offset =

(g1 – g2) x2 200 𝐿

e) Find the elevation of stations on the curve by eqn Elevation of station on curve at a distance x = elevation of station on tangent ∓ tangent offset or correction -ve sing in case of crest curve, +ve sign in case of sag curve.

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Curves

Vertical Curves Example 01: Find the length of vertical curve connecting two uniform grades from the following data:

a) + .8% and - 0.6%, rate of change of grade is 0.1 per 30m b) – 0.5% and +1%, rate of change of grade is 0.05 per 30m Solution: a) L =

(g1% − g2%) (0.8 – (−.6)) = 𝑥 30 = 420 m r% 0.1

b) L =

(−0.5 – (+ 1)) 0.05

𝑥 30 = 900 m

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Vertical Curves Example 02: Calculate the R.L of various stations pegs on a vertical curve connecting two uniform grades of + 0.5% and -0.7%. The chainage and R.L of point of intersection are 500 m and 350.750 m respectively. Take the rate of change of grade as 0.1% per 30m.

Solution: 1) Length of vertical curve L (g1% − g2%) (0.5 – (−.7)) L= x 30 = 𝑥 30 = 360 m r% 0.1 𝐿 = 180 𝑚 2 2) Chainage of A = 500 m 𝐿

Chainage of O = chainage of A - 2 = 500 – 180 = 320 m 𝐿

Chainage of B = chainage of A + 2 = 500 + 180 = 680 m 25

Solution: 02

Vertical Curves

3) R.L of Point of intersection A = 330.75 m 𝑔𝐿

0.5 (360) 200

= 329.85 m

𝑔𝐿

0.7 (360) 200

= 329.49 m

1 R.L of O = 330.75 - 200 = 330.75 2 R.L of B = 330.75 - 200 = 330.75 -

R.L of mid point E of chord OB = ½ (R.L of O + R.L of B) = ½ (329.85 +329.49) = 329.67 m R.L of F (Vertex of curve) = ½ ( R.L of B + R.L of A) = ½ (330.75 + 329.67) = 330.21 m

The difference AF b/w A and F = 330.75 – 330.21 = 0.54 m Check: AF =

𝒈𝟏 −𝒈𝟐 𝑳 𝟖𝟎𝟎

=

𝟎.𝟓 −(−𝟎.𝟕 𝟑𝟔𝟎 𝟖𝟎𝟎

= 𝟎. 𝟓𝟒 𝒎 𝑶𝒌

4) 1st point on the curve chainage = 350 m , x= 30m R.L of 1st point on tangent = 329.85 + Tangent correction =y =

(g1 – g2) x2 200 𝐿

𝑔1 𝑥 100

= 330 m

= 0.015 m

R.L of 1st point on the curve = 330 m – 0.015 = 329.985 m

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Vertical Curves Solution: 02 Station

Chainage (m)

Grade Elevation (m)

Tangent Correction (m)

Curve Elevation (m)

0

320

329.85

0

329.85

1

350

330.00

− 0.015

329.985

2

380

330.15

− 0.06

330.090

3

410

330.30

− 0.135

330.165

4

440

330.45

− 0.240

330.210

5

470

330.60

− 0.375

330.225

6 (F)

500

330.75

− 0.54

330.210

7

530

330.54

− 0.375

330.165

8

560

330.33

− 0.240

330.090

9

590

330.12

− 0.135

329.985

10

620

330.91

− 0.06

329.850

11

650

330.70

− 0.015

329.685

12 (B)

680

330.49

0

329.490

Start of V.C

Vertex of V.C

End od V.C

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Vertical Curves Assignment Three cases of Sight distance: Example 3 and 4 , page 220

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