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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.
I
θ = 60º 00’ 00”.

Radius, R

250m

Deflection angle , θ

600

Offset

20m

Chainage intersection point, I

2942 m

T1

T2

R = 250
m
Tangent length

= R tan θ/2
= 250 tan 60 °/2

= 144.34m

Chainage T 1

= chainage I – tangent length
= 2942 - 144.34

Arc length

= 2π x R x θ
360
= 2π x 250 x 60 o
360

Chainage T 2
Ofset,Y

0
20
40
60
80
100
120
140
144.338

=
=
R

250
250
250
250
250
250
250
250
250

PROCEDU
RE
X = R − R −Y
2

= 2797.66m

= 261.8m

chainage T 1 + arc length
2797.66 + 261.8
= 3059.46m
R2

Y2

62500
0
62500
400
62500
1600
62500
3600
62500
6400
62500
10000
62500
14400
62500
19600
62500 20833.333

R 2 -Y 2

62500
62100
60900
58900
56100
52500
48100
42900
41666.667

√(R 2 -Y 2)
250.000
249.199
246.779
242.693
236.854
229.129
219.317
207.123
204.124

X= R-√(R 2 Y 2)
0.000
0.801
3.221
7.307
13.146
20.871
30.683
42.877
45.876

2
Radius, R

12m

Tangent length = 12m

Deflection angle , θ

900

Chainage T 1

= 8m

Arc length

= 18.85m

Chainage T 2

= 26.85m

Offset

2m

Chainage intersection point, I

20 m

Offset, Y
0
2
4
6
8
10
12

R
12
12
12
12
12
12
12

R2
144
144
144
144
144
144
144

Y2
0
4
16
36
64
100
144

R2-Y2
144
140
128
108
80
44
0

√(R2-Y2)
12.000
11.832
11.314
10.392
8.944
6.633
0.000

X= R-√(R2-Y2)
0.000
0.168
0.686
1.608
3.056
5.367
12.000
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.
I
θ = 42º

Radius, R

320m

Deflection angle , θ

420

Offset

20m

Chainage intersection point, I

2020 m

T1

T2

R = 320
m
Long chord length

PROCEDUR
w
E

= 2R sin θ/2
= 2 x 320 sin 42°/2

114.678 m
Tangent length

= 229.355m

= R tan θ/2
= 320 tan 42°/2

= 184.752 m

Chainage T 1

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

Arc length

= 2π x R x θ
360
= 2π x 320 x 42 o
360

Chainage T 2

=
=

chainage T 1 + arc length
1835.245 + 335.103
= 2170.351 m

X = R − Y − R − (W / 2)
2

= 335.103 m

2

2

2

w/2 =
Radius, R

12m

Long chord length =
16.97m
W/2
= 8.485m

Deflection angle , θ

900

Tangent length = 12m

2m

Chainage intersection point, I

20 m

Offset, Y

R

R2

Y2

0
2
4
6
8
8.485

12
12
12
12
12
12

144
144
144
144
144
144

0
4
16
36
64
72.000

(w/2)2

R2-Y2

72.000 144
72.000 140
72.000 128
72.000 108
72.000
80
72.000 72.000

√(R2-Y2)

12.000
11.832
11.314
10.392
8.944
8.485

= 8m

Arc length

= 18.85m

Chainage T 2

Offset

Chainage T 1

= 26.85m

R2-(w/2)2 √(R2-(W/2)2

72.000
72.000
72.000
72.000
72.000
72.000

8.485
8.485
8.485
8.485
8.485
8.485

X= √(R2-Y2)√(R2 -(W/2)2)

3.515
3.347
2.828
1.907
0.459
0.000
Scale
1m :1cm
1:100

Tangent length = 12m
Offset
= 2 m

I

2)

O 12
O 10

O 0m

m

m

8m

T1

=

2

12

t
fse
of

m

16
0.

O 2m

8m

m

O 6m

m

7m
36
5.

=

m

6
05
3.

O 4m

m
2c O
1

m
08
1.6

12

m

6
68
0.

Offset, Y X= R-√(R -Y
0
0.000
2
0.168
4
0.686
6
1.608
8
3.056
10
5.367
12
12.000
2
8.845m

O8m

O6m

O4m

1.907m

W/2 = 8.845m
O2m

O0m
2m
W = 16.971 m

O2m
O4m
O6m
O8m
2m
2m
2m

2.828m

W/2 = 8.845m

3.515m

3.515m

3.515m

2.828m

1.907m

X= √(R2-Y2)Offset, Y √(R2 -(W/2)2)
3.515
0
3.347
2
2.828
4
1.907
6
0.459
8
0.000
8.485
O
T1

2.828m

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

2.828m

Scale
1m :1cm
1:100

O8.845m
T

0.459 m

2
Given data of curve ranging was as follows:Radius
Deflection angle
Offset
Chainage I

=
=
=
=

650 m
17058’50”
20m
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.
Radius, R

I

650m

θ = 17º 58’ 50”.

Given
Deflection angle , 17º 58’ 50”.
θ
Offset
20m
Chainage
intersection
point, I

δ
δ

1 (deg ree )

1 (min ute )

1718.9 x C
=
R

T2

4100m
R = 650
m

Draw the table form for
deflection angle method

formula

1718.9 x C
=
60 R

T1

Stn
.

Chainag
e

Chord
length

Deflection
angle,δ
(0 ‘ “)

Setting out
angle, δ
(0 ‘ “)
PROCEDUR
1718.9 x 2.837
δ =
E60xx650.000
1718.9 20

Tangent length

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

Chainage T 1

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

Arc length

= R x θ x 2π
360
= 650 x 17 o 58’50” x 2 π = 188.292m
360

Chainage T 2

Stn.

=
=

Chainage

δ =

60 x650
1718.9 x 5.455
δ =
chainage T
+ arc length
3997.163 + 188.292 = 4185.455m
60 x650
1

Chord length, C

Deflection angle,δ

Setting out angle, δ

T1

3997.163

0

00 0’ 0”

00 0’ 0”

δ1

4000

2.837

00 19’ 30”

00 19’ 30”

δ2

4020

20.000

00 52’ 53”

10 12’ 24”

δ3

4040

20.000

00 52’ 53”

20 5’ 17”

δ4

4060

20.000

00 52’ 53”

20 58’ 10”

δ5

4080

20.000

00 52’ 53”

30 51’ 4”

δ6

4100

20.000

00 52’ 53”

40 43’ 57”

δ7

4120

20.000

00 52’ 53”

50 36’ 50”

δ8

4140

20.000

00 52’ 53”

60 29’ 44”

δ9

4160

20.000

00 52’ 53”

70 22’ 37”

δ10

4180

20.000

00 52’ 53”

80 15’ 31”

T2

4185.455

5.455

00 37’ 30”

80 53’ 1”

Σ = 188.292

Σ = 80 53’ 1”

θ / 2 = 170 58’50” / 2 = 80 53’ 1”
Tangent length = 14.261m

Radius, R
Deflection angle , θ

60

Offset

5m

Chainage intersection point, I

Stn.

24.7m

20 m

Chainage

Chord length, C

Chainage T 1

= 5.739m

Arc length
25.866m

=

Chainage T 2
31.605m

0

=

Deflection angle,δ

Setting out angle, δ

T1

5.739

0

00 0’ 0”

00 0’ 0”

δ1

10

4.261

40 56’ 32”

40 56’ 32”

δ2

15

5.000

50 47’ 57”

100 44’ 29”

δ3

20

5.000

50 47’ 57”

160 32’ 26”

δ4

25

5.000

50 47’ 57”

220 20’ 23”

δ5

30

5.000

50 47’ 57”

280 8’ 20”

T2

31.605

1.605

10 51’ 42”

300 0’ 2”

Σ = 25.866

Σ = 300 00’ 2”

θ / 2 = 600 / 2 = 300
Scale
1m :1cm
1:100
I

1m
6
.2
14

T1

=

m
1c
6
.2
14

T2
I

T1

Scale
1m :1cm
1:100

T2
Given data of curve ranging was as follows:Radius
Deflection angle
Chainage I

=
=
=

600 m
18024’
2140m

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

PROCEDUR
1718.9 x 17.179
δ =
E60xx600.000
1718.9 20

= R tan θ/2
= 600 tan 18°24′/2 = 97.20m

Chainage T 1

= chainage I – tangent length
= 2140.00 - 97.20
= 2042.80m

Arc length

= R x θ x π
360
= 600 x 18 o 24’ x π
360

Chainage T 2

Stn.

=
=

Chainage

δ =

chainage T 1 + arc length
2042.80 + 192.68
=

Chord length, C

60 x600
1718.9 x 15.506
δ =
2235.48m
60 x600

= 192.684m

Deflection angle,δ
(0 ‘ “), T1

T1

2042.821

0

00 0’ 0”

δ1

2060

17.179

00 49’ 12”

δ2

2080

20.000

00 57’ 18”

δ3

2100

20.000

00 57’ 18”

δ4

2120

20.000

00 57’ 18”

δ5

2140

20.000

00 57’ 18”

δ6

2160

20.000

00 57’ 18”

δ7

2180

20.000

00 57’ 18”

δ8

2200

20.000

00 57’ 18”

δ9

2220

20.000

00 57’ 18”

T2

2235.506

15.506

00 44’ 25”

Σ = 192.684

Σ = 90 12’ 1”

Deflection angle,δ
(0 ‘ “), T2
360° - θ + δ 1
2
Example δ1
= 360° - θ + δ 1
2
= 360° - 18024’ +
2
= 351° 37’ 20”
Stn.

Chainage

Chord length, C

Deflection angle,δ
(0 ‘ “), T1

00 49’ 12”

Deflection angle,δ
(0 ‘ “), T2

T1

2042.821

0

00 0’ 0”

350° 48’ 00”

δ1

2060

17.179

00 49’ 12”

351° 37’ 20”

δ2

2080

20.000

00 57’ 18”

352° 34’ 40”

δ3

2100

20.000

00 57’ 18”

353° 32’ 00”

δ4

2120

20.000

00 57’ 18”

δ5

2140

20.000

00 57’ 18”

355° 26’ 20”

δ6

2160

20.000

00 57’ 18”

356° 23’ 40”

δ7

2180

20.000

00 57’ 18”

357° 21’ 00”

δ8

2200

20.000

00 57’ 18”

358° 18’ 20”

δ9

2220

20.000

00 57’ 18”

359° 15’ 40”

T2

2235.506

15.506

00 44’ 25”

360° 00’ 00”

Σ = 192.684

Σ = 90 12’ 1”

354° 29’ 20”

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Tutorial circular curve

  • 2. 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. I θ = 60º 00’ 00”. Radius, R 250m Deflection angle , θ 600 Offset 20m Chainage intersection point, I 2942 m T1 T2 R = 250 m
  • 3. Tangent length = R tan θ/2 = 250 tan 60 °/2 = 144.34m Chainage T 1 = chainage I – tangent length = 2942 - 144.34 Arc length = 2π x R x θ 360 = 2π x 250 x 60 o 360 Chainage T 2 Ofset,Y 0 20 40 60 80 100 120 140 144.338 = = R 250 250 250 250 250 250 250 250 250 PROCEDU RE X = R − R −Y 2 = 2797.66m = 261.8m chainage T 1 + arc length 2797.66 + 261.8 = 3059.46m R2 Y2 62500 0 62500 400 62500 1600 62500 3600 62500 6400 62500 10000 62500 14400 62500 19600 62500 20833.333 R 2 -Y 2 62500 62100 60900 58900 56100 52500 48100 42900 41666.667 √(R 2 -Y 2) 250.000 249.199 246.779 242.693 236.854 229.129 219.317 207.123 204.124 X= R-√(R 2 Y 2) 0.000 0.801 3.221 7.307 13.146 20.871 30.683 42.877 45.876 2
  • 4. Radius, R 12m Tangent length = 12m Deflection angle , θ 900 Chainage T 1 = 8m Arc length = 18.85m Chainage T 2 = 26.85m Offset 2m Chainage intersection point, I 20 m Offset, Y 0 2 4 6 8 10 12 R 12 12 12 12 12 12 12 R2 144 144 144 144 144 144 144 Y2 0 4 16 36 64 100 144 R2-Y2 144 140 128 108 80 44 0 √(R2-Y2) 12.000 11.832 11.314 10.392 8.944 6.633 0.000 X= R-√(R2-Y2) 0.000 0.168 0.686 1.608 3.056 5.367 12.000
  • 5. 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. I θ = 42º Radius, R 320m Deflection angle , θ 420 Offset 20m Chainage intersection point, I 2020 m T1 T2 R = 320 m
  • 6. Long chord length PROCEDUR w E = 2R sin θ/2 = 2 x 320 sin 42°/2 114.678 m Tangent length = 229.355m = R tan θ/2 = 320 tan 42°/2 = 184.752 m Chainage T 1 = chainage I – tangent length = 2020 - 134.752 = 1835.245 m Arc length = 2π x R x θ 360 = 2π x 320 x 42 o 360 Chainage T 2 = = chainage T 1 + arc length 1835.245 + 335.103 = 2170.351 m X = R − Y − R − (W / 2) 2 = 335.103 m 2 2 2 w/2 =
  • 7. Radius, R 12m Long chord length = 16.97m W/2 = 8.485m Deflection angle , θ 900 Tangent length = 12m 2m Chainage intersection point, I 20 m Offset, Y R R2 Y2 0 2 4 6 8 8.485 12 12 12 12 12 12 144 144 144 144 144 144 0 4 16 36 64 72.000 (w/2)2 R2-Y2 72.000 144 72.000 140 72.000 128 72.000 108 72.000 80 72.000 72.000 √(R2-Y2) 12.000 11.832 11.314 10.392 8.944 8.485 = 8m Arc length = 18.85m Chainage T 2 Offset Chainage T 1 = 26.85m R2-(w/2)2 √(R2-(W/2)2 72.000 72.000 72.000 72.000 72.000 72.000 8.485 8.485 8.485 8.485 8.485 8.485 X= √(R2-Y2)√(R2 -(W/2)2) 3.515 3.347 2.828 1.907 0.459 0.000
  • 8. Scale 1m :1cm 1:100 Tangent length = 12m Offset = 2 m I 2) O 12 O 10 O 0m m m 8m T1 = 2 12 t fse of m 16 0. O 2m 8m m O 6m m 7m 36 5. = m 6 05 3. O 4m m 2c O 1 m 08 1.6 12 m 6 68 0. Offset, Y X= R-√(R -Y 0 0.000 2 0.168 4 0.686 6 1.608 8 3.056 10 5.367 12 12.000 2
  • 9. 8.845m O8m O6m O4m 1.907m W/2 = 8.845m O2m O0m 2m W = 16.971 m O2m O4m O6m O8m 2m 2m 2m 2.828m W/2 = 8.845m 3.515m 3.515m 3.515m 2.828m 1.907m X= √(R2-Y2)Offset, Y √(R2 -(W/2)2) 3.515 0 3.347 2 2.828 4 1.907 6 0.459 8 0.000 8.485 O T1 2.828m Long chord length = 16.971 w/2 = 8.845m Offset = 2 m 2.828m Scale 1m :1cm 1:100 O8.845m T 0.459 m 2
  • 10. Given data of curve ranging was as follows:Radius Deflection angle Offset Chainage I = = = = 650 m 17058’50” 20m 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.
  • 11. Radius, R I 650m θ = 17º 58’ 50”. Given Deflection angle , 17º 58’ 50”. θ Offset 20m Chainage intersection point, I δ δ 1 (deg ree ) 1 (min ute ) 1718.9 x C = R T2 4100m R = 650 m Draw the table form for deflection angle method formula 1718.9 x C = 60 R T1 Stn . Chainag e Chord length Deflection angle,δ (0 ‘ “) Setting out angle, δ (0 ‘ “)
  • 12. PROCEDUR 1718.9 x 2.837 δ = E60xx650.000 1718.9 20 Tangent length = R tan θ/2 = 650 tan (17º 58’ 50”/2) = 102.837m Chainage T 1 = chainage I – tangent length = 4100.00 - 102.837 = 3997.163m Arc length = R x θ x 2π 360 = 650 x 17 o 58’50” x 2 π = 188.292m 360 Chainage T 2 Stn. = = Chainage δ = 60 x650 1718.9 x 5.455 δ = chainage T + arc length 3997.163 + 188.292 = 4185.455m 60 x650 1 Chord length, C Deflection angle,δ Setting out angle, δ T1 3997.163 0 00 0’ 0” 00 0’ 0” δ1 4000 2.837 00 19’ 30” 00 19’ 30” δ2 4020 20.000 00 52’ 53” 10 12’ 24” δ3 4040 20.000 00 52’ 53” 20 5’ 17” δ4 4060 20.000 00 52’ 53” 20 58’ 10” δ5 4080 20.000 00 52’ 53” 30 51’ 4” δ6 4100 20.000 00 52’ 53” 40 43’ 57” δ7 4120 20.000 00 52’ 53” 50 36’ 50” δ8 4140 20.000 00 52’ 53” 60 29’ 44” δ9 4160 20.000 00 52’ 53” 70 22’ 37” δ10 4180 20.000 00 52’ 53” 80 15’ 31” T2 4185.455 5.455 00 37’ 30” 80 53’ 1” Σ = 188.292 Σ = 80 53’ 1” θ / 2 = 170 58’50” / 2 = 80 53’ 1”
  • 13. Tangent length = 14.261m Radius, R Deflection angle , θ 60 Offset 5m Chainage intersection point, I Stn. 24.7m 20 m Chainage Chord length, C Chainage T 1 = 5.739m Arc length 25.866m = Chainage T 2 31.605m 0 = Deflection angle,δ Setting out angle, δ T1 5.739 0 00 0’ 0” 00 0’ 0” δ1 10 4.261 40 56’ 32” 40 56’ 32” δ2 15 5.000 50 47’ 57” 100 44’ 29” δ3 20 5.000 50 47’ 57” 160 32’ 26” δ4 25 5.000 50 47’ 57” 220 20’ 23” δ5 30 5.000 50 47’ 57” 280 8’ 20” T2 31.605 1.605 10 51’ 42” 300 0’ 2” Σ = 25.866 Σ = 300 00’ 2” θ / 2 = 600 / 2 = 300
  • 16. Given data of curve ranging was as follows:Radius Deflection angle Chainage I = = = 600 m 18024’ 2140m Based on data-data given above, •Provide a table of setting out by two theodolite without measuring tape.
  • 17. Tangent length PROCEDUR 1718.9 x 17.179 δ = E60xx600.000 1718.9 20 = R tan θ/2 = 600 tan 18°24′/2 = 97.20m Chainage T 1 = chainage I – tangent length = 2140.00 - 97.20 = 2042.80m Arc length = R x θ x π 360 = 600 x 18 o 24’ x π 360 Chainage T 2 Stn. = = Chainage δ = chainage T 1 + arc length 2042.80 + 192.68 = Chord length, C 60 x600 1718.9 x 15.506 δ = 2235.48m 60 x600 = 192.684m Deflection angle,δ (0 ‘ “), T1 T1 2042.821 0 00 0’ 0” δ1 2060 17.179 00 49’ 12” δ2 2080 20.000 00 57’ 18” δ3 2100 20.000 00 57’ 18” δ4 2120 20.000 00 57’ 18” δ5 2140 20.000 00 57’ 18” δ6 2160 20.000 00 57’ 18” δ7 2180 20.000 00 57’ 18” δ8 2200 20.000 00 57’ 18” δ9 2220 20.000 00 57’ 18” T2 2235.506 15.506 00 44’ 25” Σ = 192.684 Σ = 90 12’ 1” Deflection angle,δ (0 ‘ “), T2
  • 18. 360° - θ + δ 1 2 Example δ1 = 360° - θ + δ 1 2 = 360° - 18024’ + 2 = 351° 37’ 20” Stn. Chainage Chord length, C Deflection angle,δ (0 ‘ “), T1 00 49’ 12” Deflection angle,δ (0 ‘ “), T2 T1 2042.821 0 00 0’ 0” 350° 48’ 00” δ1 2060 17.179 00 49’ 12” 351° 37’ 20” δ2 2080 20.000 00 57’ 18” 352° 34’ 40” δ3 2100 20.000 00 57’ 18” 353° 32’ 00” δ4 2120 20.000 00 57’ 18” δ5 2140 20.000 00 57’ 18” 355° 26’ 20” δ6 2160 20.000 00 57’ 18” 356° 23’ 40” δ7 2180 20.000 00 57’ 18” 357° 21’ 00” δ8 2200 20.000 00 57’ 18” 358° 18’ 20” δ9 2220 20.000 00 57’ 18” 359° 15’ 40” T2 2235.506 15.506 00 44’ 25” 360° 00’ 00” Σ = 192.684 Σ = 90 12’ 1” 354° 29’ 20”