2. Highway and railroad routes are chosen after studying all possible
locations. This involves;
Use of aerial imagery, satellite imagery and ground surveys as well as
analysis of existing plans and maps. Design requirement with minimal
social, environmental and financial impact should be considered.
{
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12. 3 Methods of Defining
Degree of Curvature
1. Radius of curvature; (often in highways) R is
selected as a multiple of 100 ft. The smaller R, the
sharper the curve.
2. Degree of curvature on the chord basis is the
central angle subtended by a chord of 100 ft.
3. Degree of curvature on the arc basis is the central
angle of a circle subtended by an arc of 100 ft.
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13. R&
Chord basis 100 ft arc
100 ft chord
50 ft 50 ft R
R
Arc basis
/2
50 5729.58
R R
sin
13 2
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14. 100 ft R&
100 – ft arc
R
Sharp curve (large , small R)
R
14
Flat curve (small , large R)
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15. R&
Degree of Curvature & Radius of Curvature
R, Chord basis R, Arc basis
= 1 5729.65 ft 5729.58 ft
= 5 1146.28 ft 1145.92 ft
= 10 573.69 ft 572.96 ft
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16. L
R R
360◦
c = 2R
Relationship between the degree of curve () and the circle
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18. The degree of curve () is used as a
term to define the sharpness of the
curve.
( )
{is that central angle subtended
by 100 ft of arc.
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19. and R
From previous figure,
/360 = 100/2R
= 5729.58/R
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20. Arc
From previous figure,
L/100 = θ/
L = 100 (θ/)
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21. Problem
For a horizontal circular curve, these data are
given; the P.I. is at station 64 + 32.2, is 24 20
and an of 4 00 has been selected. Compute
the necessary data and set up the field notes for
50-ft stations.
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22. Computation of Necessary Data
5729.58 5729.58
R 1432.39 ft
4
T R tan (1432.39) (0.21560) 308.82 ft
2
L.C. 2 R sin (2)(1432.39) (0.21076) 603.78 ft
2
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23. Computation of Necessary Data
1 1
E R[ 1] (1432.39) ( 1) 32.91 ft
cos 12 10
cos
2
M R (1 cos ) (1432.39) (1 cos 1210) 32.17 ft
2
100 (100) (24.333333)
L 608.33 ft
4
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25. As the degree of curvature is between 3 and 7,
the curve will be staked out with 50-ft chords. The
distance from B.C. (61 + 23.4) to the first 50-ft
station (61 + 50) is 26.4 ft and the deflection angle
to be used for that point is (26.6/100) (/2) =
(26.6/100) (4/2) = 0 31 55. For each of
subsequent 50-ft stations from 61 + 50 to 67 + 00,
the deflection angles will increase by /4 or 1 00
00. Finally, for the E.C. at station 67 + 31.7, the
deflection angle will be 11 32 + (31.7/100) (4/2)
= 12 10 00 = /2 as it should.
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26. Field Notes
Station Points Deflection angles Curve data
67 + 31.7 E.C. 12 10 00
67 + 00 11 31 55
…………………. …………………….
64 + 00 5 31 55 P.I. = 64 + 32.2
+ 50 4 31 55 = 24 20
63 + 00 3 31 55 = 4 00
+ 50 2 31 55 R = 1432.4 ft
62 + 00 1 31 55 T = 308.8 ft
+ 50 0 31 55 E = 32.9 ft
+ 23.4 B.C. 0 00 L = 608.3 ft
61 + 00
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27. Deflection
Angles
EC
BC
1 2
Length of curve, chord basis
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28. Deflection
Angles
• 3 Use 100 - ft Chords.
• = 3 - 7 Use 50 - ft Chords.
• = 7 - 14 Use 25 - ft Chords.
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29. Arcs & Chords
Comparison
R = 5729.58/ = 2 = 3 = 6 = 10
Chord (ft) C = 2 R sin (deflec. angle) 99.995 99.989 99.954 99.873
Arc (ft) 100 100 100 100
For flat curves, say = 2/ 3 chord arc
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30. Deflection Angles
[(25/100) (3/2)] = 0 22.5
[0 22.5 + (3/2)] = 1 52.5
BC
Back tangent
= 3
Staking out a horizontal curve
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32. Chord Calculations
C = 2R (sin deflection angle)
R = 400.000 m
First chord C = 2 400 (sin 0 14 01) = 3.2618 m = 3.262 m
(at three decimals, chord = arc)
Even station chord C = 2 400 (sin 1 25 57) = 19.998 m
Last chord C = 2 400 (sin 0 27 42) = 6.448 m
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33. Metric Considerations
The full station is 1 km (e.g. 1 + 000), and the
stakes are at 50-m, 20-m and 10-m intervals.
In field work, the use of (as opposed to R) allows
quick determination of the deflection angle for even
stations.
In metric system, would be the central angle
subtended by 100 m of arc and the deflection
angles would be similarly computed as have done
for ft units.
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34. Metric Considerations
If = 6, the deflection angles would be as follows:
Cross stations (m) Deflection angles
100 /2 = 300
50 /4 = 130
20 /5 = 036
10 /10 = 018
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35. Field Procedures to Lay
Out A Curve
• Extend the two straight lines 1 and 2, to determine
the P.I. point of intersection
• Measure (Intersection angle).
• Select .
• Two or more of these elements should be known or
assumed , R, T, E or .
• Calculate the stations B.C. = P.I. – T
& E.C. = B.C. + L
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36. Horizontal Curves Passing Through Certain Points
T
T-x x P.I.
y 35 10
C
A T
R
R-y
/2
/2
R
36
B
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37. P.I.
TS
TS
Circular curve
T.S. S.T.
C.S.
S.C.
LS LS
R
R
Spiral Curves
1st spiral curve 2nd spiral
curve
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38. S.P.I. S.C.
Circular curve
T.S.
R
Enlargement of Spiral Curve
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39. Moving up on PI
θ = 12 51
the Curve
0 + 240
3 06
EC
Tree
BC
6 12
θ = 12 51
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40. Problem of Offset Curves
Given Data:
= 12 51, R = 400.000 m and PI at 0 + 241.782
Calculated Data:
T = 45.044 m, L = 89.710 m, BC at 0 + 196.738 and EC at 0 + 286.448
Required:
Curbs to be laid out on 6-m offsets at 20-m stations.
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41. PI
LS
PI ¢
PI EC
RS
BC
θ = 12 51
O
41
Offset Curves
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