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24 HORIZONTAL CURVES
Asterisks indicate problems that have partial answers given in Appendix G.
24.1 How is the effect of centrifugal force counteracted on a horizontal curve?
24.2 What is the difference between a curve defined by the arc definition and one defined by
the chord definition?
24.3 For the following circular curves having a radius R, what is their degree of curvature by
(1) arc definition and (2) chord definition?
Problems 24.4 through 24.6. Use the chord definition for the railroad curve and the arc definition
for the highway curves.
(*)
L
600
476.92
196.350
T
304.53
239.93
99.456
E
32.01
16.25
9.796
M
31.31
16.10
9.607
LC
595.74
475.47
195.090
PC
33+40.47
22+25.07
6+418.044
PTBack
42+45.00
29+41.92
6+713.850
PTForward
39+49.53
27+04.93
6+616.956
Tabulate R or D, T, L, E, M, PC, PT, deflection angles, and incremental chords to lay out
the circular curves at full stations (100 ft or 30 m) in Problems 24.7 through 24.14.
24.7 Highway curve with Da = 2°30′, I = 10°30′, and PI station = 36 +44.50 ft.
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24.10 Highway curve with R = 900 m, I = 12°30′, and PI station = 4+200.600 m.
Station | Chord | Defl. Increment | Defl. Angle |
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24.11 Highway curve with R = 1200 ft, I = 30°00′, and PI station = 45 + 50.00 ft.
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24.12 Highway curve with L = 270 m, R = 600 m, and PI station = 4 + 350.000 m.
Station | Chord | Defl. Increment | Defl. Angle |
4+212.675
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24.13 Highway curve with T = 131.65 ft, R = 1200 ft, and PI station = 67 + 50.00 ft.
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all
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24.14 Railroad curve with T = 300.00 ft, DC = 2°30′, and PI station = 48 + 00.00 ft.
In Problems 24.15 through 24.18 tabulate the curve data, deflection angles, and total
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24.15 The curve of Problem 24.7.
24.16 The curve of Problem 24.8
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24.17 The curve of Problem 24.9
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24.18 The curve of Problem 24.10
24.19 A rail line on the center of a 80-ft street makes a 55°24′ turn into another street of equal
width. The corner curb line has R = 10 ft. What is the
largest R that can be given a circular curve for the track
centerline if the law requires it to be at least 5 ft from the
24.20 A highway survey PI falls in a pond, so a cut off line AB = 275.21 ft is run between the
tangents. In the triangle formed by points A, B, and PI, the angle at A = 16°28′ and at B =
22°16′. The station of A is
54+92.30 ft. Calculate and
tabulate curve notes to run, by
deflection angles and
incremental chords, a 4°30′
(arc definition) circular curve
at full-station increments to
connect the tangents.
55°24'
A
E5'
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copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.
24.21 In the figure, a single
circular highway curve
(arc definition) will join
tangents XV and VY and
also be tangent to BC.
Calculate R, L, and the
stations of the PC and PT.
sin 55 56
PI = 2500 + 231.79 = 27 + 31.79
PC = 2731.79 583.67 =
21 + 48.12
PT 2148.12 + 1073.13 = 32 + 21.25
PT = 2731.79 + 583.67 = 33 + 15.46
Back
Forward
24.22* Compute Rx to fit requirements of the figure and
make the tangent distances of the two curves equal.
875 1392.04 ft
tan 64 18 16.5
x
x
R
24.23 After a backsight on the PC with set on the instrument, what is the deflection angle
to the following circular curve points?
0 00
1750 ft
R = 550 ft RX
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24.24 In surveying a construction alignment, why should the I angle be measured using both
faces of the instrument?
24.25 A highway curve (arc definition) to the right, having R = 500 m and I = 15°30′, will be
laid out by coordinates with a total station instrument setup at the PI. The PI station is
3 + 855.200 m, and its coordinates are X = 75,428.863 m and Y = 36,007.434 m. The
azimuth (from north) of the back tangent proceeding toward the PI is 45°00′00″. To
orient the total station, a backsight will be made on a POT on the back tangent. Compute
lengths and azimuths necessary to stake the curve at 30-m stations.
Station
δ
Total δ
Chord
Chord Azimuth
3+922.416
1°17'04"
7°45'00"
134.851
52°45'1"
3+900.000
1°43'08"
6°27'56"
112.608
51°27'57"
3+870.000
1°43'08"
4°44'48"
82.752
49°44'49"
3+840.000
1°43'08"
3°01'40"
52.822
48°1'41"
3+810.000
1°18'33"
1°18'33"
22.845
46°18'33"
3+787.153
45°0'0"
24.26 In Problem 24.25, compute the XY coordinates at 30-m stations.
Station
Azimuth
Chord
X
Y
3+922.416
52°45'01"
134.851
75,488.089
36,040.941
3+900.000
51°27'57"
112.608
75,468.833
36,029.470
3+870.000
49°44'49"
82.752
75,443.903
36,012.789
3+840.000
48°01'41"
52.822
75,420.018
35,994.643
3+810.000
46°18'33"
22.845
75,397.265
35,975.098
3+787.153
45°00'00"
75,380.746
35,959.317
24.27 An exercise track must consist of two semicircles and two tangents, and be exactly 1000
m along its centerline. The two tangent sections are 200 m each. Calculate the radius for
the curves.
What sight distance is available if there is an obstruction on a radial line through the PI
inside the curves in Problems 24.28 and 24.29?
24.29 For Problem 24.9, obstacle 10 m from curve.
24.30 If the misclosure for the curve of Problem 24.7, computed as described in Section 24.8,
is 0.05 ft, what is the field layout precision?
24.31 Assume that a 100-ft entry spiral will be used with the curve of Problem 24.7. Compute
and tabulate curve notes to stake out the alignment from the TS to ST at full stations
using a total station and the deflection-angle, total chord method.
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all
24.32 Same as Problem 24.31, except use a 200-ft spiral for the curve of Problem 24.8.
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24.33 Same as Problem 24.31, except for the curve of Problem 24.10, with a 50-m entry spiral
using stationing of 30 m and a total station instrument.
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**************************************
24.34 Compute the area bounded by the two arcs and tangent in Problem 24.22.
Area = 1699285 − 305,414 - 1,087,408 = 306,460
24.35 In an as-built survey, the XY coordinates in meters of three points on the centerline of a
highway curve are determined to be A: (3770.52, 4913.84); B: (3580.80, 4876.37); C:
(3399.27, 4809.35). What are the radius, and coordinates for the center of the curve in
meters?
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24.36 In Problem 24.35, if the (x, y) coordinates in meters of two points on the centerline of
the tangents are (3042.28, 4616.77) and (4435.66, 4911.19), what are the coordinates of
the PC, PT, and the curve parameters L, T, and I?
L = 638.23
T = 326.59
"2.03'26230
57.1209
28.524
tan
31.131862.370119.491138.391166.44352
"43'28316360
15.915
10.869
tan
07.126262.370177.461638.391128.30421
1
2
22
1
1
22
O
O
Az
O
Az
O
Solve triangle O-1-PC:
70.32138.122007.1262122 PC
∠1-O-PC = acos(1220.38/122.07) = 14°46ʹ04ʺ
Solve triangle O-2-PT:
61.49838.122031.1318222 PT
∠PT-O-2 = acos(1220.38/1318.31) = 22°13ʹ25ʺ
AzO1 = 316°28’43” + 14°46’04” = 331°14’47”
AzO2 = 23°26ʹ03ʺ – 22°13ʹ25ʺ = 1°12ʹ38ʺ
XPC = 3911.38 + 1220.38 sin(331°14ʹ47ʺ) = 3324.32
YPC = 3701.62 + 1220.38 cos(331°14ʹ47ʺ) = 4771.52
XPT = 3911.38 + 1220.38 sin(1°12ʹ38ʺ) = 3937.16
YPT = 3701.62 + 1220.38 cos(1°12ʹ38ʺ) = 4921.72
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