Erbil Polytechnic University · Vertical Curves. Assumptions and Terminology In general, vertical curves are designed such that the two tangent lengths (grades) are equal. The length

1Survey II

Lecture 8

Out lines

Curves.

Vertical Curves.

References.

2

3Vertical Curves.

Curves are needed to provide smooth transitions between straight

segments (tangents) of grade lines for highways and railroads.

Because these curves exist in vertical planes, they are called

vertical curves.

4Vertical Curves.

Vertical curves are used to connect intersecting straights in thevertical plane.

These straights are usually referred to as gradients (grades) and thecombination of the gradients and vertical curve is knows as thevertical alignment.

Gradients are usually expressed as percentages and their maximumvalues should not exceed 8%.

Purposes of vertical curves is to ensure:

– passengers in vehicles travelling along the curves are

transported safely and comfortably.

– there is adequate visibility to enable vehicles to be able

either to stop or to overtake safely.

5Vertical Curves.

The vertical curves can either be crest curves or sag curves.

A crest curve, which can also be referred to as a summit or

hogging curve, is one for which the algebraic difference of the

gradients is negative.

A sag curve, which can also be referred to as a valley or sagging

curve, is one for which the algebraic difference of the gradients

is positive.

𝐴 = 𝐴𝑙𝑔𝑒𝑏𝑟𝑎𝑡𝑖𝑐 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑔𝑟𝑎𝑑𝑒𝑠 = 𝑔2 − 𝑔1

6Vertical Curves.

There are six different combinations of grades:

Crest Curve ( A, r,Δe, andΔy are negative).

Sag Curve (A, r, Δe, and Δy are Positive).

7Vertical Curves.

Assumptions and Terminology

In general, vertical curves are designed such that the two tangent

lengths (grades) are equal.

The length is the same whether measured along the tangents, the

chord, the horizontal or the curve itself.

Terminology:

L = Length of Vertical Curve measured horizontally (e.g. L=350

m = 3.5 station).

g1 , g2 = Percent longitudinal grades or slopes (with sign).

8Vertical Curves.

Assumptions and Terminology

A = Algebraic difference in grades=g2‐g1

r= Rate of change of grade per station.

P.V.C = Point of Vertical Curvature.

P.V.I= Point of Vertical Intersection.

P.V.T= Point of Vertical Tangency.

Δy = Difference in elevation between tangent and curve.

Δe= Difference in elevation at P.V.I

y= Elevation of a Point on curve.

9Vertical Curves.

Assumptions and Terminology

X= Horizontal distance in station from P.V.C to required point.

X˳= Horizontal distance in station from P.V.C to the highest

(crest) or the lowest (sag) point.

y˳= Elevation of the highest (crest) or the lowest (sag) point.

10Vertical Curves.

11General equation of a vertical parabolic curve

𝑦𝑝 = 𝑎 + 𝑏𝑋𝑝 + 𝑐𝑋𝑝2

𝑎 = Elev. of PVC

𝑏 = 𝑔1𝑋

𝑐𝐿2 = 𝑔1

𝐿

2+ 𝑔

2

𝐿

2− 𝑔

1𝐿

𝑐 = (𝑔2−𝑔1

2𝐿)

𝑟 =𝑔2− 𝑔1

𝐿

𝑐 =𝑟

2

12

𝑦 = 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶 + 𝑔1𝑋 +𝑟

2𝑋2

𝐴 = 𝑔2− 𝑔1

𝑟 =𝐴

𝐿=𝑔2 − 𝑔1

𝐿

∆𝑦 =𝑟

2∗ 𝑥2

∆𝑒 =𝑟

2∗ 𝑥2 =

𝐴

2𝐿

𝐿

2

2

=𝐴𝐿

8

∆𝑒 =

𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶 + 𝐸𝑙𝑒𝑣. 𝑃𝑉𝑇2

− 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐼

2

𝑦 =𝑟

2𝑥2 + 𝑔1𝑥 + 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶

𝑆𝑡𝑎. 𝑃𝑉𝐼 = 𝑆𝑡𝑎. 𝑃𝑉𝐶 +𝐿

2

General equation of a vertical parabolic curve

13General equation of a vertical parabolic curve

𝑆𝑡𝑎. 𝑃𝑉𝑇 = 𝑆𝑡𝑎. 𝑃𝑉𝐼 +𝐿

2

𝐸𝑙𝑒𝑣. 𝑃𝑉𝐼 = 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶+g1

𝐿

2

𝐸𝑙𝑒𝑣. 𝑃𝑉T=𝐸𝑙𝑒𝑣.𝑃𝑉I+g2

𝐿

2

𝑥𝑜 = −𝑔1

𝑟

𝑦𝑜 =𝑟

2𝑥𝑜

2 + 𝑔1 𝑥𝑜 + 𝐸𝑙𝑒𝑣. 𝑜𝑓 𝑃𝑉𝐶

14Vertical Curve Computations

𝑆𝑡𝑒𝑝𝑠:1. Compute the stations and elevations of main points (PVC, PVI,and PVT).2. Compute (A and r).3. Compute X for required point = Sta. of a point‐ Sta. of PVC.4. Compute elevation by parabolic equation.5. Tabulate the results as follows:

15Vertical Curve Computations

Station X(sta.) 𝒚 =𝒓

𝟐𝑿𝟐 + 𝒈𝟏𝑿 + 𝒆𝒍𝒆𝒗. 𝒐𝒇 𝑷𝑽𝑪

PVC 0+0 Elev. PVC

----------------- X1 𝒚 =𝒓

𝟐𝑿𝟏𝟐 + 𝒈𝟏𝑿 + 𝒆𝒍𝒆𝒗. 𝒐𝒇 𝑷𝑽𝑪

------------------ X2 𝒚 =𝒓

𝟐𝑿𝟐𝟐 + 𝒈𝟏𝑿 + 𝒆𝒍𝒆𝒗. 𝒐𝒇 𝑷𝑽𝑪

PVI 𝐿

2𝒚 =

𝒓

𝟐(𝑳

𝟐)𝟐+𝒈𝟏𝑿 + 𝒆𝒍𝒆𝒗. 𝒐𝒇 𝑷𝑽𝑪

----------------- X3 𝒚 =𝒓

𝟐𝑿𝟑𝟐 + 𝒈𝟏𝑿 + 𝒆𝒍𝒆𝒗. 𝒐𝒇 𝑷𝑽𝑪

------------------ X4 𝒚 =𝒓

𝟐𝑿𝟒𝟐 + 𝒈𝟏𝑿 + 𝒆𝒍𝒆𝒗. 𝒐𝒇 𝑷𝑽𝑪

PVT L Elev. of PVT

16𝐸𝑥𝑎𝑚𝑝𝑙𝑒:−For the following informations ∶

g1=+1.4%, g2=-1.1%, Elev. of PVI = 153.55m St. of PVI = 11+34.45, and

Curve length = 135.50m

Find:-

Elev. and station of PVC, PVT and points on curve at 20m interval.

Solution:-

PVC Elev. = PVI Elev. – g1

𝐿

2= 153.55 -

1.4

100𝑥135.5

2= 152.60m

PVT Elev. = PVI Elev. – g2

𝐿

2= 153.55 –

1.1

100𝑥135.5

2= 152.80 m

PVC St. = PVI St. -𝐿

2= 1134.45 -

135.5

2= 10+66.70

PVT St. = PVI St. +𝐿

2= 1134.45 +

135.5

2= 12+02.20

𝑟 =𝐴

𝐿=

𝑔2−𝑔

1

𝐿=

−1.1%−1.4%

135.50=

−1.1

100−1.4

100

135.50= - 0.0001845

17

PVT

PVI

PVC

18𝐸𝑥𝑎𝑚𝑝𝑙𝑒:−

𝑦 =𝑟

2𝑥2 + 𝑔1𝑥 + 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶 = 𝑦 =

−0.0001845

2𝑥2 + 1.5%𝑥 + 152.60

No. Station 𝒙 𝒓

𝟐. 𝒙𝟐 𝒈𝟏𝒙 Elev. m

PCV 10+66.70 152.60

10+80 13.3 - 0.01632 +0.1862 152.76

3 11+00 33.3 -0.10230 +0.4662 152.96

4 11+20 53.3 -0.26210 +0.7462 153.08

5 11+40 73.3 -0.49565 +1.0262 153.13

6 11+60 93.3 -0.80302 +1.3062 153.10

11+80 113.3 -1.18420 +1.5862 153.00

12+00 133.3 -1.63918 +1.8662 152.83

PVT 12+02.20 152.80

References 19

Ghilani, C. D. and P. R. WOLF (2014). Elementary Surveying: An Introduction to Geomatics . New Jersey, PEARSON.

Uren, J. and B. Price (2010). Surveying for Engineers. UK, PALGRAVE MACMILLAN.

Barry F. Kavanagh – 7th – ed. SURVEYING with ConstructionApplications. PEARSON.

20End of Lecture 8