1. Circular curves

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

14. Scale
1m :1cm
1:100
I
1m
6
.2
14
T1
=
m
1c
6
.2
14
T2

15. I
T1
Scale
1m :1cm
1:100
T2

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”