Circular curves

One circular curve of radius 250 meter will build up for connected two straights road. The chainage of intersection point, I being 2942 meter and the deflection angle being 60º 00’ 00”. The curve will be mark at every offset of 20 meter. Calculate the setting out data required to peg the curve with offset method from tangent line.

Radius, R 250m

Deflection angle , 600

Offset 20m

Chainage intersection point, I

2942 m

I

T1

T2

= 60º 00’ 00”.

R = 250 m

PROCEDUPROCEDURE RE Tangent length = R tan θ/2

= 250 tan 60/2 = 144.34m

Chainage T1 = chainage I – tangent length= 2942 - 144.34 = 2797.66m

Arc length = 2 x R x 360= 2 x 250 x 60o = 261.8m 360

Chainage T2 = chainage T1 + arc length = 2797.66 + 261.8 = 3059.46m

Ofset,Y R R2 Y2 R2-Y2

√(R2-Y2)

X= R-√(R2-Y2)

22 YRRX

0 250 62500 0 62500 250.000 0.00020 250 62500 400 62100 249.199 0.80140 250 62500 1600 60900 246.779 3.22160 250 62500 3600 58900 242.693 7.30780 250 62500 6400 56100 236.854 13.146100 250 62500 10000 52500 229.129 20.871120 250 62500 14400 48100 219.317 30.683140 250 62500 19600 42900 207.123 42.877

144.338 250 62500 20833.333 41666.667 204.124 45.876

Radius, R 12m

Deflection angle , 900

Offset 2m

Chainage intersection point, I

20 m

Offset, Y R R2 Y2 R2-Y2 √(R2-Y2)

X= R-√(R2-Y2)

0 12 144 0 144 12.000 0.0002 12 144 4 140 11.832 0.1684 12 144 16 128 11.314 0.6866 12 144 36 108 10.392 1.6088 12 144 64 80 8.944 3.05610 12 144 100 44 6.633 5.36712 12 144 144 0 0.000 12.000

Tangent length = 12m

Chainage T1 = 8m

Arc length = 18.85m

Chainage T2 = 26.85m

The centre-line of two straights is projected forward to meet at I, the deflection angle being 42°. If the straights are to be connected by a circular curve of radius 320 m, tabulate all the setting-out data, assuming 20-m chords on a through chainage basis, the chainage of I being 2020 m. Calculate the setting out data required to peg the curve with offset method from long chord line.

Radius, R 320m

Deflection angle , 420

Offset 20m

Chainage intersection point, I 2020 m

I

T1

T2

= 42º

R = 320 m

Long chord length = 2R sin θ/2

= 2 x 320 sin 42/2 w = 229.355m w/2 = 114.678 m

Tangent length = R tan θ/2 = 320 tan 42/2 = 184.752 m

Chainage T1 = chainage I – tangent length= 2020 - 134.752 = 1835.245 m

Arc length = 2 x R x 360= 2 x 320 x 42o = 335.103 m 360

Chainage T2 = chainage T1 + arc length = 1835.245 + 335.103 = 2170.351 m

PROCEDURPROCEDURE E

22 )2/(22 WRYRX

Radius, R 12m

Deflection angle , 900

Offset 2m

Chainage intersection point, I

20 m

Long chord length = 16.97mW/2 = 8.485m

Tangent length = 12m

Chainage T1 = 8m

Arc length = 18.85m

Chainage T2 = 26.85m

Offset, Y R R2 Y2 (w/2)2 R2-Y2 √(R2-Y2) R2-(w/2)2 √(R2-(W/2)2

X= √(R2-Y2)- √(R2 -(W/2)2)

0 12 144 0 72.000 144 12.000 72.000 8.485 3.5152 12 144 4 72.000 140 11.832 72.000 8.485 3.3474 12 144 16 72.000 128 11.314 72.000 8.485 2.8286 12 144 36 72.000 108 10.392 72.000 8.485 1.9078 12 144 64 72.000 80 8.944 72.000 8.485 0.459

8.485 12 144 72.000 72.000 72.000 8.485 72.000 8.485 0.000

Offset, Y

X= R-√(R2-Y2)

0 0.0002 0.1684 0.6866 1.6088 3.05610 5.36712 12.000

Tangent length = 12mOffset = 2 m

I

12 m

=

12cm

T1

offse

t = 2

m

O 0m

O 2m

O 4m

O 6m

O 8m

O 10m

O 12m

0.168m

0.686m

1.608m

3.056m

5.367m 12 m

Scale1m :1cm1:100

Offset, Y

X= √(R2-Y2)- √(R2 -(W/2)2)

0 3.5152 3.3474 2.8286 1.9078 0.459

8.485 0.000

Long chord length = 16.971w/2 = 8.845m Offset = 2 m

Scale1m :1cm1:100

W = 16.971 m

T1

T2

W/2 = 8.845m W/2 = 8.845m

2 m

2 m

2 m

2 m

0.459 m O0m

O2m

O4m

O6m

O8m

O8.845m

O2m

O4m

O6m

O8m

O8.845m

3.5

15

m

3.5

15

m

2

.82

8m

1

.907

m

2

.82

8m

3.5

15

m

2

.82

8m

1.9

07

m

2

.82

8m

Given data of curve ranging was as follows:-

Radius = 650 mDeflection angle = 17058’50”Offset = 20mChainage I = 4100m

Based on data-data given above, •Sketch the position of the circular curve.•Provide a table of setting out by one theodolite & one measuring tape.

Given Radius, R 650m

Deflection angle , 17º 58’ 50”.

Offset 20m

Chainage intersection point, I

4100m

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 ‘ “)

I

T1

T2

= 17º 58’ 50”.

R = 650 m

Tangent length = R tan θ/2 = 650 tan (17º 58’ 50”/2) = 102.837m

Chainage T1 = chainage I – tangent length= 4100.00 - 102.837 = 3997.163m

Arc length = R x x 2 360= 650 x 17o58’50” x 2 = 188.292m 360

Chainage T2 = chainage T1 + arc length = 3997.163 + 188.292 = 4185.455m

PROCEDURE PROCEDURE

Stn. Chainage Chord length, C

Deflection angle, Setting out angle,

T1 3997.163 0 00 0’ 0” 00 0’ 0”

1 4000 2.837 00 19’ 30” 00 19’ 30”

2 4020 20.000 00 52’ 53” 10 12’ 24”

3 4040 20.000 00 52’ 53” 20 5’ 17”

4 4060 20.000 00 52’ 53” 20 58’ 10”

5 4080 20.000 00 52’ 53” 30 51’ 4”

6 4100 20.000 00 52’ 53” 40 43’ 57”

7 4120 20.000 00 52’ 53” 50 36’ 50”

8 4140 20.000 00 52’ 53” 60 29’ 44”

9 4160 20.000 00 52’ 53” 70 22’ 37”

10 4180 20.000 00 52’ 53” 80 15’ 31”

T2 4185.455 5.455 00 37’ 30” 80 53’ 1”

= 188.292 = 80 53’ 1” θ / 2 = 170 58’50” / 2 = 80 53’ 1”

65060837.29.1718

xx

65060000.209.1718

xx

65060455.59.1718

xx

Radius, R 24.7m

Deflection angle , 600

Offset 5m

Chainage intersection point, I

20 m

Tangent length = 14.261m

Chainage T1 = 5.739m

Arc length = 25.866m

Chainage T2 = 31.605m

Stn. Chainage Chord length, C

Deflection angle, Setting out angle,

T1 5.739 0 00 0’ 0” 00 0’ 0”

1 10 4.261 40 56’ 32” 40 56’ 32”

2 15 5.000 50 47’ 57” 100 44’ 29”

3 20 5.000 50 47’ 57” 160 32’ 26”

4 25 5.000 50 47’ 57” 220 20’ 23”

5 30 5.000 50 47’ 57” 280 8’ 20”

T2 31.605 1.605 10 51’ 42” 300 0’ 2”

= 25.866 = 300 00’ 2” θ / 2 = 600 / 2 = 300

I

14.2

61 m =

14.261cm

T1

Scale1m :1cm1:100

T2

I

T1

Scale1m :1cm1:100

T2

Given data of curve ranging was as follows:-

Radius = 600 mDeflection angle = 18024’Chainage I = 2140m

Based on data-data given above, •Provide a table of setting out by two theodolite without measuring tape.

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 360= 600 x 18o24’ x = 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 ‘ “), T1

Deflection angle,(0 ‘ “), T2

T1 2042.821 0 00 0’ 0”

1 2060 17.179 00 49’ 12”

2 2080 20.000 00 57’ 18”

3 2100 20.000 00 57’ 18”

4 2120 20.000 00 57’ 18”

5 2140 20.000 00 57’ 18”

6 2160 20.000 00 57’ 18”

7 2180 20.000 00 57’ 18”

8 2200 20.000 00 57’ 18”

9 2220 20.000 00 57’ 18”

T2 2235.506 15.506 00 44’ 25”

= 192.684 = 90 12’ 1”

60060179.179.1718

xx

60060000.209.1718

xx

60060506.159.1718

xx

360 - + 1

2

Stn. Chainage Chord length, C

Deflection angle,(0 ‘ “), T1

Deflection angle,(0 ‘ “), T2

T1 2042.821 0 00 0’ 0” 350 48’ 00”

1 2060 17.179 00 49’ 12” 351 37’ 20”

2 2080 20.000 00 57’ 18” 352 34’ 40”

3 2100 20.000 00 57’ 18” 353 32’ 00”

4 2120 20.000 00 57’ 18” 354 29’ 20”

5 2140 20.000 00 57’ 18” 355 26’ 20”

6 2160 20.000 00 57’ 18” 356 23’ 40”

7 2180 20.000 00 57’ 18” 357 21’ 00”

8 2200 20.000 00 57’ 18” 358 18’ 20”

9 2220 20.000 00 57’ 18” 359 15’ 40”

T2 2235.506 15.506 00 44’ 25” 360 00’ 00”

= 192.684 = 90 12’ 1”

Example 1= 360 - + 1

2= 360 - 18024’ + 00 49’ 12”

2= 351 37’ 20”