Location horizontal and vertical curves Theory

Setting out of works
horizontal and vertical curves
Horizontal Alignment
ØAn introduction to horizontal curve &Vertical curve.
ØTypes of curves.
ØElements of horizontal circular curve.
ØGeometric of circular curve
Ø Methods of setting out circular curve
Ø Setting out of horizontal curve on ground
Ø Vertical curve Definition.
ØElements of the vertical curves.
ØAvailable methods for computing the elements of vertical curves
Types of Curves
1- Horizontal Curves
2- Vertical Curves
Horizontal Curves
are circular curves. They connect tangent lines around
obstacles, such as building, swamps, lakes, change
direction in rural areas, and intersections in urban areas.
-Compound Curve.
-Reverse curve.
-Transition or Spiral Curves.
-Horizontal Curve: Simple circular
curve
-Elements of horizontal curves.
-Formulas for simple circular
curves.
-Properties of circular curves.
example:A horizontal curve having R= 500m, ∆=40°, station P.I=
12+00 ,prepare a setting out table to set out the curve
using deflection angle from the tangent and chord length
method, dividing the arc into 50m stations.
:Example H.W
A Horizontal curve is designed with a 600m radius and is
known to have a tangent of 52 m the PI is Station
200+00 determent the Stationing of the PT?
-PROCEDURE SETTING OUT Practical .
Vertical Curves
Elevation and Stations of main points on the Vertical Curve .
Assumptions of vertical curve projection.
Example: A vertical parabola curve 400m long is to be set
between 2% (upgrade) and 1% (down grade), which meet
at chainage of 2000 m, the R.L of point of intersection of
the two gradients being (500.00 m). Calculate the R.L of
the tangent and at every (50m) parabola.
Thank you all
Prepared by:
Asst. Prof. Salar K.Hussein
Mr. Kamal Y.Abdullah
Asst.Lecturer. Dilveen H. Omar
Erbil Polytechnic University
Technical Engineering College
Civil Engineering Department

1. 1
Erbil Polytechnic University
Technical Engineering College Civil Engineering Department
2021-2022
Horizontal and Vertical Curve

2. Horizontal Alignment
ØAn introduction to horizontal curve &Vertical curve.
ØTypes of curves.
ØElements of horizontal circular curve.
ØGeometric of circular curve
ØMethods of setting out circular curve
ØSetting out of horizontal curve on ground
ØVertical curve Definition.
ØElements of the vertical curves.
ØAvailable methods for computing the elements of vertical
curves
2

3. Types of Curves
1-Horizontal Curves
2-Vertical Curves
HorizontalCurves
arecircularcurves.Theyconnecttangentlinesaround
obstacles,suchasbuilding,swamps,lakes,change
directioninruralareas,andintersectionsinurbanareas.
3

4. Types of Horizontal Curves
SimplecircularcurveItisacurveofsingleconstant
radiusconnectingtwostraights.
4

5. Compound Curve
Whentherearemorethanoneradiusofanarcconnectingtwo
intersectingstraightsiscalledcompoundcurvesoftworadiusor
morejoinedtogether.
5

6. Reverse curve
Twocirculararcstangenttoeachother,withtheircenters
onoppositesidesofthealignment.
6

7. Transition or Spiral Curves:-
—When circular curve is introduced between straights it
becomes very difficult for a fast moving vehicle to change
from straight linepath having zerocurvature to a curved
path having a specified curvature therefore transition
curves are provided in which the curvature varies
gradually from zero of the straight line path to specific
value depending upon the radius of the circular curve
7

8. 8

9. Horizontal Curve: Simple circular
curve
—Considerations for Horizontal Curves
§Safety.
§Economic evaluation
Simple Curves have 3 variables
—Radius
—Design Speed
—Super elevation
9

10. Point of Intersection (PI):the point at which the two tangents to
the curve intersect
Point of Curvature (PC):the beginning point
of the curve
Point of Tangency (PT):the end point of the curve
Delta Angle:the angle between the tangents
is also equal to the angle at the center of
the curve
Tangent Distance (T):the distance from the PC to PI or from the PI
to PT
10

11. External Distance (E):the distance from the PI to the middle
point of the curve
Middle Ordinate (M):the distance from
the middle point of the curve to the middle
of the chord joining the PC and PT
Long Chord (LC):the straight distance
Along the line joining the PC and the PT
Length of Curve(L):the difference in stationing along the
curve between the PC and the PT
11
Elements of Horizontal curve

12. Formulas for simple circular
curves
12
2
tan
D
=RT
Tangent length
D=RL
180
p
2
sin2
D
=RLc
Length of curve
Length of chord
External distance
÷
÷
ø
ö
ç
ç
è
æ
-
D
= 1
2cos
1
RE
Middle ordinate

13. 13

14. α(degree)=(90/π) * (C/R)
Chainage:itisastationwithequaldistance,oritisanequalchord.
10m,l=Lengthofsmallchord.
For200m
200/10=20chainage
205=20+5chainage
207=20+7chainage
Thestationofintersection
pointsarealwaysknown. 14
D = 2 α

15. •
15

16. 16
Example
A horizontal curve having R= 500m, ∆=40°, station P.I=
12+00 ,prepare a setting out table to set out the curve
using deflection angle from the tangent and chord length
method, dividing the arc into 50m stations.

17. 17
Solution
1-Find the length of curve = R=500 dalta= 40
L = 349.07 m
2-Find tangent length
T= R tan ∆/2 = 500 * tan 40/2 =181.99m
=181.99/50= 3.6398 =3+31.99 Chains
3-Find the station of point of curve (PC)
PC=PI -T= 12*50 -T
P.C= 600 –T ; P.C= 600-181.99 = 418.01 m = 8+18.01chains
OR
P.C. = 12 –T = (12+00) –(3+31.99) = (8 + 18.01) Chains

18. 18
4-Find point of tangent (P.T).
P.T= Station P.C+L= 418.01+349.07= 767.08m=15+17.08
5-find the individual tangential angle for the first station.
We should calculate the partial chords (C)
Here, the chainageof arc is 50m
For full chord length of 50 m (α) in degrees=(90/π)*(50/500)
= 2.865°= 2°51’ 54”
Station of PC=8+18.01; First station on the curve= 9+00
Hence, C=(9+00)-(8+18.01)=31.99m, (8+50)-(8+18.01)=31.99
For chord 50m α= 2°51’ 54”;for 18.01m chord α1=1°49´59”
OR
α(degree)=(90/π) * (C/R)
α1(degree)=90/π(31.99/500)= 1°49´59”

19. 19
Station Arc Length(m)individual tangential angle
(degree)
Cumulative
tangential angle
8+18.01
9+00
10+00
11+00
12+00
13+00
14+00
15+00
15+17.08
31.00
50
50
50
50
50
50
17.08
1°49´59˝
2°51´53˝
2°51´53˝
2°51´53˝
2°51´53˝
2°51´53˝
2°51´53˝
0°58´43˝
1°49´59˝
4°41´52˝
7°33´45˝
10°25´38˝
13°17´31˝
16°09´24˝
19°01´17˝
20°00´00˝
L=Check ∑349.07Check ∑20°00´00˝
For other stations l= 50m and α=2°51’ 54”;
For last station l=(15+17.08) –(15+00)= 17.08m hence α= 0°58’ 43”
Table for setting out the curve using deflection angle method
+=

20. :Example H.W
AHorizontalcurveisdesignedwitha600mradiusandis
knowntohaveatangentof52mthePIisStation
200+00determenttheStationingofthePT?
20

21. PROCEDURE SETTING OUT Practical
I 70m
P
C
P
T
GIVEN DEFLECTION
ANGLE , θ=13018’ 32”
13016’ 00”
21

22. PROCEDURE SETTING OUT Practical
I
P
C P
T
10 11’ 37”
20 23’ 14”
25 m25 m25
m
25
m14.37
m
30 34’ 51”
40 46’ 28”
50 58’ 5”
60 39’ 15”
22

23. Vertical Curves
Vertical curves are in the shape of a parabola. A curve in the
longitudinal section of a roadway to provide easy change of
gradient.
23

24. Vertical curves (VC) are used to connect intersecting
gradients in the vertical plane. Thus, in route design they
are provided at all changes of gradient. They should be
sufficientlylarge curvature to provide comfort to the
driver, that is, they should have a low ‘rate change of
grade’. In addition, they should afford adequate ‘sight
distances’ for safe stopping at a given design speed.

25. The type of curve generally used to connect the
intersecting gradients g1 and g2 in the simple
parabola. Its uses as a sag or crest curve
Gradient:
In vertical curve design the gradients are
expressed as percentages, with a negative for a
downgradeand a positive for an upgrade,
e.g. A downgrade of 1 in 20= 5 in 100= −5%=−g1%
An upgrade of 1 in 25=4 in 100= +4%=+g2%

26. The angle of deflection of the two intersecting
gradients is called the grade angle. The grade
angle simply represents the change of grade
through which the vertical curve deflects and is
the algebraic difference of the two gradients:
A %=( g1%−g2%)
In the above example A %=( −5%−4%) =−9%
(negative indicates a sag curve).

27. Sag and Crest Curves.

28. 1) Summit (Crest) vertical curve
The slope percentage is +X% to –Y%
Parabolic Curve

29. Parabolic Curve
¨

30. ¨

31. Parabolic Curve
¨

32. ¨

33. ¨

34. ¨

35. •r= rate of change of grade per station
•g1= initial grade in percent
•g2= final grade in percent
•L= Length of the curve in stations
•A= Algebraic difference in grade
•A= g2-g1
35

36. ELEMENTS OF VERTICAL CURVES
1 -B.V.C or P.V.C: Beginning of vertical curve or point of vertical
curvature.
2 -P.V.I: Point of vertical intersection.
3 -P.V.T or E.V.C: End of vertical curve or point of vertical
tangency.
36

37. Elevation and Stations of main points on the
Vertical Curve
•If the Station and Elevation of P.V.I is known
37

38. Assumptions of vertical curve projection:
1. L = Lcbecause the slope is very small.
2. T= L/2 since MA= MH
AH= 2AM
ØUsually the Reduced level at P.I is known and
the chainage also is known.

39. ¨

40. ¨

41. ¨

42. ¨

43. ¨

44. Example: A vertical parabola curve 400m long is to be set
between 2% (upgrade) and 1% (down grade), which meet
at chainage of 2000 m, the R.L of point of intersection of
the two gradients being (500.00 m). Calculate the R.L of
the tangent and at every (50m) parabola.

45. ¨

46. ¨

47. Distance (x)Distance (m)R.L (m)Slope (g) %
X1504972
X21004982
X31504992
X42005002
X5250499.51
X63004991
X7350498.51
X84004981

48. ¨

49. StationChainageR.L. on
straight
Off-set
y
R.L. on
curve
Remark
T18004960496.0Find the
staff
reading
of H.I.
118504970.094496.906
219004980.375497.625
319504990.844498.15
A20005001.5498.5

50. Thank you all
50