topic 4 curve lesson 2
TRANSCRIPT
Circular curvesCircular curves
Horizontal curvesHorizontal curves
DefinitionDefinition
Horizontal, circular or simple curves are curves of constant radius required to connect two straights set out on the ground.
UsageUsage
roads, railways, kerb lines, pipe lines and may be set out in several ways, depending on their length and radius.
Circular Curve GeometryCircular Curve Geometry
FORMULAFORMULA
2tan
RT
2sin2
RC
2cos1
RM
1
2secRE
3602
RL
R
Cxree 60
9.1718)(deg1
R
Cxute
9.1718)(min1
Chainage T1 = Chainage I + tangent Chainage I + tangent lengthlengthChainage T2 = Chainage Chainage T1 – arc length – arc length
Setting out methodSetting out method
Offset From The Tangent Line
Offset From The Long Chord Line
Deflection Angle MethodSub Chords Line Method
OFFSET FROM TANGENT LINE
Given Radius, R 600m
Deflection angle ,
18024'
Offset 20m
Chainage intersection point, I
2140m
Draw the table from the formula:Draw the table from the formula:
Ofset
R R2 Y2 R2-Y2
formula 22 YRRX
22 YR 22 YRRX
IP ) Tangent line Y2 X2
X1
Y1
T1
circular arc T2
PROCEDURE PROCEDURE Tangent length = R tan θ/2
= 600 tan (1824/2) = 97.20m
Chainage T1 = chainage I – tangent length= 2140.00 - 97.20 = 2042.80m
Arc length = 2 x R x 360= 2 x 600 x 18o24’ = 192.68m 360
Chainage T2 = chainage T1 + arc length = 2042.80 + 192.68 = 2235.48m
Ofset R R2 Y2 R2-Y2
(1) (2) (3) = (2)2 (4) = (1)2 (5) = (3) – (4) (6) = √(5) (7) = (2) – (6)
0
6006002 = 360000
02 = 0360000 – 0 =
360000√ 360000 =
600.000600 – 600 =
0.000
20 202 = 40360000 – 40=
359600√ 359600 =
599.667600 – 599.667 =
0.333
40 402 = 1600360000 – 1600 =
358400√ 3584000 =
598.665600 – 598.665 =
1.335
60 3600 356400 596.992 3.008
80 6400 353600 594.643 5.357
97.20 9447.84 350969.109 592.426 7.574
22 YR 22 YRRX
OFFSET FROM LONG CHORD LINE
Given Radius, R 600m
Deflection angle ,
18024'
Offset 20m
Chainage intersection point, I
2140m
Draw the table from the formula:Draw the table from the formula:
Ofset R R2 Y2 R2-Y2
formula
22 YR 22 )2/(WR
22 )2/(22 WRYRX
22 )2/(22 WRYRX
PROCEDURE PROCEDURE Long chord length = 2R sin θ/2
= 2 x 600 sin (1824/2) w = 191.857m w/2 = 95.929 m
Tangent length = R tan θ/2 = 600 tan (1824/2) = 97.20m
Chainage T1 = chainage I – tangent length= 2140.00 - 97.20 = 2042.80m
Arc length = 2 x R x 360= 2 x 600 x 18o24’ = 192.68m 360
Chainage T2 = chainage T1 + arc length = 2042.80 + 192.68 = 2235.48m
Ofset
R R2 Y2 R2-Y2 w/2 2 x
(1) (2) (3) = (2)2 (4) = (1)2 (5) = (3) – (4) (6) = √(5) (7) (8) = √(3) – (7) (9) =(6) – (8)
0
6006002 = 360000
02 = 0360000 -0 =
360000√ 360000 =
600.000
95.9292 = 9202.277
√(360000 – 9202.277)=
592.282
600 - 692.282 = 7.785
20 202 = 40360000 – 40 =
359600√ 3596000 =
599.667
599.667 – 592.282 =
7.385
40 402 = 1600 358400 598.665 6.383
60 3600 356400 596.992 4.710
80 6400 353600 594.643 2.361
95.929 9202.277 350797.723 592.282 0.000
22 YR 22 )2/(wR
DEFLECTION ANGLE DEFLECTION ANGLE METHODMETHOD
Given
Radius, R 600m
Deflection angle ,
18024'
Offset 20m
Chainage intersection point, I
2140m
R
Cxree 60
9.1718)(deg1
formula
RCx
ute
9.1718)(min1
Draw the table form Draw the table form for deflection angle for deflection angle methodmethod
Stn.
Chainage
Chord length
Deflection angle,(0 ‘ “)
Setting out angle, (0 ‘ “)
Tangent length = R tan θ/2 = 600 tan (1824/2)= 97.20m
Chainage T1 = chainage I – tangent length= 2140.00 - 97.20 = 2042.80m
Arc length = R x x 2 360= 600 x 18o24’ x 2 = 192.684m 360
Chainage T2 = chainage T1 + arc length = 2042.80 + 192.68= 2235.48m
PROCEDURE PROCEDURE
Stn. Chainage Chord length, C
Deflection angle,(0 ‘ “) – use formula
Setting out angle, (0 ‘ “) – cumulative deflection angle
T1 2042.821 0 00 0’ 0” 00 0’ 0”
1 2060 17.179 00 49’ 12” 00 49’ 12”
2 2080 20.000 00 57’ 18” 10 46’ 30”
3 2100 20.000 00 57’ 18” 20 43’ 48”
4 2120 20.000 00 57’ 18” 30 41’ 6”
5 2140 20.000 00 57’ 18” 40 38’ 24”
6 2160 20.000 00 57’ 18” 50 35’ 42”
7 2180 20.000 00 57’ 18” 60 33’ 0”
8 2200 20.000 00 57’ 18” 70 30’ 18”
9 2220 20.000 00 57’ 18” 80 27’ 36”
T2 2235.506 15.506 00 44’ 25” 90 12’ 1”
= 192.684 = 90 12’ 1” θ / 2 = 180 24’ / 2 = 90 12’ 1”
6006017.1799.1718
xx
60060000.209.1718
xx
60060506.159.1718
xx
+=
+=
Sub chords line Sub chords line methodmethod
• Given Radius, R 600m
Deflection angle ,
18024'
Offset 20m
Chainage intersection point, I
2140m
formula
Draw the table form Draw the table form for sub chord line for sub chord line methodmethodStn.
Chainage
Chord length
Offset
Ra
Ofset2
2
1
Rabb
Ofset2
)(2
Rb
Ofsetn
2
Rbcc
Ofsetn 2
)(1
Tangent length = R tan θ/2 = 600 tan (1824/2)= 97.20m
Chainage T1 = chainage I – tangent length= 2140.00 - 97.20 = 2042.80m
Arc length = R x x 2 360= 600 x 18o24’ x 2 = 192.684m 360
Chainage T2 = chainage T1 + arc length = 2042.80 + 192.68= 2235.48m
PROCEDURE PROCEDURE
Stn. Chainage Chord length
Offset
T1 2042.821 0 -1 2060 a = 17.179 0.2462 2080 b = 20.000 0.6203 2100 20.000 0.6674 2120 20.000 0.6675 2140 20.000 0.6676 2160 20.000 0.667
7 2180 20.000 0.6678 2200 20.000 0.6679 2220 20.000 0.667T2 2235.506 c =15.506 0.459
= 192.684 -
6002179.17 2
1 xOfset
6002)179.1720(20
2 xOfset
600202
n
Ofset
6002)20506.15(506.15
1 xOfset
n
First OffsetSecond Offset
Other Offset
Last Offset
PROCEDURE SETTING PROCEDURE SETTING OUT OUT
Circular curves may be set out in a variety of ways, depending on the accuracy required, its radius of curvature and obstructions on site.
Methods of setting out are as follows:
•Using one theodolite and a tape by the tangent angle method. This method can be used on all curves, but is necessary for long curves of radius unless they are set out by coordinates.
•Using two theodolites. This method can be used on smaller curves where the whole length is visible from both tangent points and where two instruments are available.
•Using tapes only by the method of offsets from the tangent. This method is used for minor curves only.
•Using tapes only by the method of offsets from the long chord. This method is used for short radius curves.
OPTICAL SQUAREOptical squares are simple sighting instruments used to set out right angles. They can be provided either with mirrors or with one or two prisms. Because of practical difficulties in using squares with mirrors, they have been replaced by squares with prisms: "prismatic squares". There are two major types of prismatic squares: single prismatic squares and double prismatic squares; both will be dealt with in the sections which follow.
FIELD WORKFIELD WORK
OFFSET LINE FROM BASELINE USED OFFSET LINE FROM BASELINE USED OPTICAL SQUAREOPTICAL SQUARE
STEP 1STEP 1STEP 2STEP 2
OFFSET LINE FROM BASELINEOFFSET LINE FROM BASELINE
STEP 1STEP 1 STEP 2STEP 2
STATION AT BASELINESTATION AT BASELINE
End of sub -topic• ExerciseExercise
• Practical 6 & 7 Practical 6 & 7