2. introduction
HORIZONTAL CURVES
As a highway changes horizontal direction, turning to
change the vehicle direction at the point of intersection
between the two straight lines is not feasible. The change in
direction would be too abrupt and too risky for the safety of
modern, high-speed vehicles, the driver and its passengers. It
is therefore necessary to interpose a curve between the
straight lines. Horizontal curves occur at locations where
two roadways intersect, providing a gradual transition
between the two. The straight lines of a road are called
tangents because the lines are tangent to the curves used to
change direction.
kaila marie joy d.r. turla
4. Simple Curves
The simple curve is an arc of a circle. The radius
of the circle determines the sharpness or flatness
of the curve.
PI
I
T E
M
PC PT
C
R
I
kaila marie joy d.r. turla
5. Elements of a Simple Curve
CIRCULAR CURVE ELEMENTS
R = radius of the curve
T = tangent distance
I = intersection of
deflection angle
E = external distance
M = middle ordinate
LC = long chord or line
connecting PC and PT
PC = point of curvature or
beginning point
PT = point of tangency or
end of the curve
D = degree of curve
L = length of curve
kaila marie joy d.r. turla
9. Sample Problem:
The angle of intersection of a
circular curve is 36030'. Compute R,
T, sta PC and PT, and the Degree
of Curve if the external distance is PI
12.02m, point of intersection is at
sta 75+040. Use 10 m per station. T E
M
PC PT
Given:
I= 36030'
R
E= 12.02m
Sta PI= 75+040 I
kaila marie joy d.r. turla
10. Solution:
R = E/(sec I/2 - 1)
= 12.02 . = 226.942m
sec (36 30'/2) -1
T = R (tan I/2)
= 226.942 tan (36 30'/2) = 74.83m
Sta PC= Sta V - T
= (75+040) - 74.83 = 74 + 965.17
D = 10(360) . = 10(360) . = 2031'
2(3.1416)R 2(3.1416)(226.94)
LC = 10I = 10 (36031') = 145.03 m
D 2031'
Sta PT= Sta PC + LC = (74+965.17) + 145.03 = 75 + 110.20
kaila marie joy d.r. turla
12. Laying Out of Simple Curves
INSTRUMENTS AND ACCESSORIES:
as reference for the different abbreviations and
terminologies used in this exercise.
Theodolite or Transit
Steel Tape 3. All values needed to lay out the curve should be
Chaining Pins tabulated accord
Stakes or Hubs
4. Set up and level the instrument at the designated
vertex or point or intersection (PI).
PROCEDURE:
1. Before proceeding to the designated survey site, the 5. Establish on the ground the PC by laying out with a
lab instructor should be consulted with respect to the steel tape the computed tangent distance (T) from the PI.
following curve elements which will be needed to define the The intersection angle (I) at the PI and the distance
circular curve to be laid out: carried through the forward tangent will also be needed
a. Radius of the curve (R) to set a stake at the PT.
b. Intersection or deflection angle (I)
c. Stationing of the point of intersection (PI) 6. Transfer and set up the instrument at the PC. At the
PC lay off the total deflection angle from PI to PT and
2. Similarly, the different elements of the circular check if the stake previously set up at the PT is along the
curve such as: T, L, LC, E, M, and the stationings of the PC line of sight. If it doesn't check, an error exists in either
and PT should be determined by calculations. The measurement or computation. As an added check, stake out
computations should also include the deflection angles and the midpoint of the curve before beginning to set
chord lengths which will be needed when staking out the intermediate stations. By intersecting the angle (180-I) at
curve by half section intervals. The accompanying sketch is the PI and laying off the external distance (E), the midpoint
given to serve can be established. A check of the deflection angle from the
PC to the midpoint should equal to I/4.
kaila marie joy d.r. turla
13. 7.. To establish the first curve station, first set the
horizontal circle reading of the instrument to zero and
sight along the back tangent. Then turn the instrument
about its vertical axis and lay off the required sub-
deflection angle and the corresponding chord distance for
the first station. Set a hub to mark the located station.
8. With the first station already established, now lay
out the next chord length from it, and locate the second
station on the intersection of the line of sight (defined by
the next deflection angle) and the end of the chord. Also
set a hub to mark this located station. CIRCULAR CURVE ELEMENTS
9. Repeat the process of locating stations on the curve
R = radius of the curve
by laying out the computed deflection angles and the chord
distances from the previously established station. Do this T = tangent distance
until all the required stations of the curve are laid out I = intersection of deflection angle
and properly marked on the ground.
E = external distance
10. When the final station is established, the closing PT M = middle ordinate
should be staked out using the final deflection angle and LC = long chord or line connecting PC and PT
subchord, to determine the misclosure in laying out the
PC = point of curvature or beginning point
curve.
PT = point of tangency or end of the curve
D = degree of curve
L = length of curve
kaila marie joy d.r. turla
14. Sample Problem:
It is required to lay-out a
simple curve by deflection
angles. The curve is to
connect two tangents with
an angle of 300 and radius
of 130m. Compute the
tangent distance, length of
curve, deflection angles to
each station of the curve.
Sta PC is at 5 + 767.20.
} Radii
Tangent
LC
Deflection Angles
Chord Distance
kaila marie joy d.r. turla
15. Solution:
T = R (tan I/2)
= 130 tan (30/2) = 34.83m
D = 1145.916 . = 1145.916 . = 8049'
R 130
LC = 20I = 20 (300) . = 68.07 m
D 80 49'
Sta PT= Sta PC + LC = (5 + 767.20) + 68.07 = 5 + 835.27
20 = 68.07 = 12.8 = 15.27 B1= 8o49'
B1 300 B2 B3 B2= 5o28'
B3= 6o49'
Cn = 2R(sin I/2 )
C1 = 2(130)(sin 20 49') = 12.78
C2 = 2R(sin 70 13') = 32.66
C3 = 2R(sin 110 38') = 52.43
C4 = 2R(sin 150 ) = 67.29
kaila marie joy d.r. turla
18. Compound Curves
Frequently, the terrain will require the use of the compound curve. This curve
normally consists of two simple curves joined together and curving in the
same direction.
I1 + I 2
T
I1 I2
T1 + T2
T1 T2
R1
R1 R2
kaila marie joy d.r. turla
19. Elements of Compound Curve
R = radius of the curve
T = tangent distance
I = intersection of
deflection angle
E = external distance
M = middle ordinate
LC = long chord or line
connecting PC and PT
PC = point of curvature or
beginning point
PT = point of tangency or
end of the curve
D = degree of curve
L = length of curve
PCC= Point of Compound
Curvature
kaila marie joy d.r. turla
21. Sample Problem:
A long chord from PC to PT of a compound
curve is 180m long and the angle it makes
with the longer tangents are 12o and 18o
respectively. Find the radius of the
compound curve if the common tangent is
parallel to the long chord.
kaila marie joy d.r. turla
22. Solution:
30 o
12 o
18 o
c1 165
o
c2
165 o
6o
12 o 6o
180 m 9o
180 m 18 o
9o
18 o
R1
R2
12 o
kaila marie joy d.r. turla
23. Solution:
180m . = ___C1____ = ____C2____
sin 1650 sin 90 sin 60
Cn = 2R(sin I/2 )
C1 = 2R1(sin I/2 )
108. 79 = 2 R1(sin I2/2 )
R1= 520.38 m
C2 = 2R2(sin I/2 )
72.70= 2R2(sin 18/2)
R2= 232.37 m
kaila marie joy d.r. turla
26. Sample Problem:
A 2.5 km roadway centerline is
established during a preliminary survey. The
three tangent section are to be connected by
two simple curves with radius of 213.5 m and
the second with a radius of 182.9 m.
Determine the station of the PC's and PT's,
the total length of the curve and the last
station of the final route.
kaila marie joy d.r. turla
27. 0 + 400
V2
R1 =213.5m
S2
PC1
PT1
S1
V1 R1 =182.9 m
35 o
0 + 500 Last Sta. 2 + 500
m
700
kaila marie joy d.r. turla
28. T1 = R1 (tan I/2) Sta PC2= Sta PT1 + S1
= 213.5 tan (35/2) = ( 0 + 563.11 ) + 585.38
T1 = 67.32 m Sta PC2 = 1 + 148.49
Sta PC1= Sta V1 - T1 D2 = 1145.916 = 1145.916 . = 6015''
= ( 0 + 500 ) - 67.32 R2 182. 40
Sta PC1 = 0 + 432.68
D = 1145.916 = 1145.916 . = 5022' LC2 = 20I2 = 20 (290) = 92.8m
R 213.5 D2 6015'
LC1 = 20I1 = 20 (350) = 130.43 Sta PT2= Sta PC2 + LC2
m = ( 1 + 148.49) + 92.8
0
D1 5 22' Sta PT2 = 1 + 241.29
Sta PT1= Sta PC1 + LC1 S2 = 1300 - T2
= ( 0 + 432.68 ) + 130.43 = 1300 - 47.30
Sta PT1 = 0 + 563.11 = 1252.70 m
S1 = 700 - (T1+T2) Last Station = Sta. PT2 + S2
T2 = R2 (tan I/2) = ( 1 + 241.29 ) + 1252.70
= 182.90 tan (29/2) Last Station = 2 + 493.99
therefore,
T2 = 47.30 m
Total Length = 2493.99m
S1 = 700 - (67.32 + 47.30 )
S1 =585. 38m
kaila marie joy d.r. turla
30. Reversed Curve
A reverse curve consists of two simple curves
joined together, but curving in opposite directions.
The curves are connected at the Point of Reversed
Curve, PRC, which is the PC of the first curve and
also the PT of the succeeding curve.
For safety reasons, this curve is seldom used in
highway construction as it would tend to send an
automobile off the road.
kaila marie joy d.r. turla
33. F
O D
S E
E S
P R S
Y E E
T V V
E R
R U
C
kaila marie joy d.r. turla
34. Since the tangents are parallel, the deflection angles are also equal.
kaila marie joy d.r. turla
35. From a common point, the backward and forward tangents lead in
different directions.
kaila marie joy d.r. turla
36. The backward and forward tangents meet or join at a particular point
called point of convergence.
kaila marie joy d.r. turla
37. An intermediate tangent lies at the common in between the PT of the
preceding curve and the PC of the next curve.
kaila marie joy d.r. turla
38. Sample Problem:
The perpendicular distance between two
parallel tangents of a reversed curve is 35m.
The azimuth of the common tangent is 300o.
If the radius of the first curve is 150m,
determine the radius of the second curve.
kaila marie joy d.r. turla
39. Solution:
a y
R2
35m
R1
b
30o
x
kaila marie joy d.r. turla
40. Solution:
R1 = a + x ; a = R 1- x
R2 = b + y ; b = R2 - y
35 = a + b
35 = (R1- x) + (R2 - y)
cos 30o = X/150 ; x= 150 cos 30o
cos 30o = y/R2 ; y= R2 cos 30o
35 = (150 - 150 cos 30o) + (R2 - R2 cos 30o)
solving for R2 R2 = 111.24m
kaila marie joy d.r. turla
