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Home»Highway Engineering»How to Determine Setting-out Data for Vertical Curves in Road Construction Works
Highway Engineering

How to Determine Setting-out Data for Vertical Curves in Road Construction Works

Mezie EthelbertBy Mezie EthelbertUpdated:
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Reading Time: 4 minutes

Vertical curves are one of the three important curves encountered in road construction works. Other important curves are the horizontal curves and the transition curves. Vertical curves are introduced in road construction works where there is change of grade from upward grade to downward grade. The vertical curve can be crest/summit curve or sag/valley curve. Summit curves are used at high points while sag curves are used at low points. The essence of vertical curves is to ensure that transition from downward grade to upward grade in the case of valley curve is changed from steep to smooth/easier. The design length of the vertical curve depends on the grade angles and the sight distances such as overtaking sight distance and stopping sight distance.

To determine the setting-out data for a vertical curve, the following information must be known;
(1) The design length of the vertical curve (this depends on the code applicable in each country). Based on Nigeria Highway Manual Part 1: Design (Geometric Design), the length of vertical curve to satisfy the requirements of minimum sight distance, comfort and appearance should not be shorter that L = KA, where L is the length of the vertical curve in metres (m), A is the algebraic difference of the gradients in percent (%). The value of K depends on the applicable design speeds and can be obtained from the Table 1 below.

(2) The gradients of the intersecting slopes and the reduced level of at least one known point or reduced level of sufficient points for the gradients to be calculated (see Figures 1 and 2).

Figure 1; Parameters considered to determine length of vertical curve considering stopping sight distance
Figure 2; Parameters considered to determine length of vertical curve considering passing sight distance

Example
Calculate the setting-out data by 25-m chainage length of a 300-m summit curve where two gradients +2% and -1.6% meet. The reduced level of the start of the curve at A is 27.94 m.

Figure 3: Representation of the Example

Solution
The rise from A to P (see Figure 3) in 150 m is calculated as follows;
2% of 150 m = (2/100) x 150 = 0.02 x 150 = + 3m (indicating rise)

The fall from P to B (see Figure 3) in 150 m is calculated as follows;
1.6% of 150 m = (1.6/100) x 150 = 0.016 x 150 = -2.4 m (indicating fall)

Reduced level (R.L.) of A (given) = 27.94 m

R.L. of P = R.L. (A) + 3 = 27.94 + 3 = 30.94 m

R.L. of B = R.L. (P) – 2.4 = 30.94 – 2.4 = 28.54 m

R.L. of C = [(R.L.(A)) + (R.L.(B))]/2 = (27.94 + 28.54)/2 = 28.24 m

Elevation, e = [(R.L.(P)) – (R.L.(C))]/2 = (30.94 + 28.24)/2 = 2.7/2 = 1.35

The vertical ordinate from the tangent is given by;
y = e (z/l)2 where ‘l’ is distance along the tangent and ‘z’ is distance along the chord length, all in metres.

Since the length of rise of the curve is 150 m and the chainage length is 25 m along the tangent, the value of ‘l’ = 150/25 = 6.
Thus, y = 1.35 (z/6)2 = 0.0375 z2

How to calculate high point details
The high point is frequently required to be pegged in order to arrange the road drainage

Chainage of the high point, X = gl/ (g – gˡ) = where g is the rising gradient (+2%) and gˡ is the falling gradient (-1.6%).
X = (2 x 300)/ (2 – (-1.6)) = 600/3.6 = 166.667 m from A or 133.333 m from B

To calculate y for the high point, y = 1.35 (166.667/150)2 = 1.35 (1.1111)2 = 1.667 m

Reduced level at the high point, AP = R.L.of A + (length x gˡ) = 27.94 + (1.667 x 2) = 27.94 + 3.334 = 31.274 m
R.L. on curve = 31.274 – 1.667 = 29.607

See Table 2 for the summary of the results and the explanation of the columns below

Note:

Column (1) records the running chainage through the curve.

Column (2) records the number of chords used as units of measure.

Columns (3) and (4) records the calculation of ‘y’ from the formula using the factor ‘0.0375’ calculated previously. The factor ‘0.0375’ may also be calculated from the grade angles as follows, (g – gˡ)/ 2n, where ‘n’ is the number of chords. ‘g’ represents the rising grade angle of +2% per chord length = (2 x 25)/ 100 = +0.5 m per chord. ‘gˡ’ represents the falling grade angle of -1.6% per chord length = (1.6 x 25)/ 100 = -0.4 m per chord. Thus, (0.5 – (- 0.4))/ (2 x 12) = 0.9/24 = 0.0375.

Column (5) records the reduced level along the tangent at each chord length by adding the rise per chord (+0.5 m) at each rising tangent up to P, then subtracting the fall per chord (-0.40 m) on the falling tangent down to B.

Column (6) records the reduced levels of the curve obtained by subtracting column (4) from column (5).

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Mezie Ethelbert

An inquisitive engineer with considerable skills in analysis, design and research in the field of civil engineering.

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