978-0134604657 Chapter 24

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.

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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

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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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

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

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”

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.

24.32 Same as Problem 24.31, except use a 200-ft spiral for the curve of Problem 24.8.

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?

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

22







































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