tutorial curve

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 , θ 60 0
T1
I
θ = 60º 00’ 00”.
T2
Offset
Chainage intersection point, I
20m
2942 m
R = 250
m

Tangent length = R tan θ/2
= 250 tan 60°/2 = 144.34m
Chainage T 1
Arc length
PROCEDU
RE
= chainage I – tangent length
= 2942 - 144.34 = 2797.66m
= 2π x R x θ
360
= 2π x 250 x 60 o = 261.8m
360
Chainage T 2
= chainage T 1
+ arc length
= 2797.66 + 261.8 = 3059.46m
Ofset,Y R R 2 Y 2 R 2 -Y 2 √(R 2 -Y 2) X= R-√(R 2 -
X
=
R
−
R
2
− Y
Y 2) 2
0 250 62500 0 62500 250.000 0.000
20 250 62500 400 62100 249.199 0.801
40 250 62500 1600 60900 246.779 3.221
60 250 62500 3600 58900 242.693 7.307
80 250 62500 6400 56100 236.854 13.146
100 250 62500 10000 52500 229.129 20.871
120 250 62500 14400 48100 219.317 30.683
140 250 62500 19600 42900 207.123 42.877
144.338 250 62500 20833.333 41666.667 204.124 45.876

Radius, R
12m
Deflection angle , θ 90 0
Tangent length = 12m
Chainage T 1
= 8m
Offset
Chainage intersection point, I
2m
20 m
Arc length = 18.85m
Chainage T 2
= 26.85m
Offset, Y R R 2 Y 2 R 2 -Y 2 √(R 2 -Y 2) X= R-√(R 2 -Y 2)
0 12 144 0 144 12.000 0.000
2 12 144 4 140 11.832 0.168
4 12 144 16 128 11.314 0.686
6 12 144 36 108 10.392 1.608
8 12 144 64 80 8.944 3.056
10 12 144 100 44 6.633 5.367
12 12 144 144 0 0.000 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.
Radius, R
Deflection angle , θ 42 0
320m
T1
I
θ = 42º
T2
Offset
Chainage intersection point, I
20m
2020 m
R = 320
m

Long chord length = 2R sin θ/2
114.678 m
= 2 x 320 sin 42°/2 w = 229.355m w/2 =
Tangent length = R tan θ/2
= 320 tan 42°/2 = 184.752 m
PROCEDUR
E
Chainage T 1
Arc length
= chainage I – tangent length
= 2020 - 134.752 = 1835.245 m
= 2π x R x θ
360
= 2π x 320 x 42 o = 335.103 m
360
Chainage T 2
= chainage T 1
+ arc length
= 1835.245 + 335.103 = 2170.351 m
X
=
R
2
− Y
2
−
R
2
)
−
( W
2
/ 2

Radius, R
12m
Deflection angle , θ 90 0
Long chord length =
16.97m
W/2 = 8.485m
Tangent length = 12m
Offset
Chainage intersection point, I
2m
20 m
Chainage T 1
= 8m
Arc length = 18.85m
Chainage T 2
= 26.85m
Offset, Y R R 2 Y 2 (w/2) 2 R 2 -Y 2 √(R 2 -Y 2) R 2 -(w/2) 2 √(R2-(W/2)2
X= √(R 2 -Y 2 )-
√(R 2 -(W/2) 2 )
0 12 144 0 72.000 144 12.000 72.000 8.485 3.515
2 12 144 4 72.000 140 11.832 72.000 8.485 3.347
4 12 144 16 72.000 128 11.314 72.000 8.485 2.828
6 12 144 36 72.000 108 10.392 72.000 8.485 1.907
8 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

Tangent length = 12m
Offset
= 2 m
Scale
1m :1cm
1:100
Offset, Y X= R-√(R 2 -Y 2)
0 0.000
2 0.168
4 0.686
6 1.608
8 3.056
10 5.367
12 12.000
T1
0.168m
0.686m
1.608m
3.056m
12 m = 12cm
O 2m
O 4m O O 8m 6m O10m O 12m
O 0m
offset = 2 m
I
5.367m
12 m

Long chord length = 16.971
w/2 = 8.845m
Offset
= 2 m
Scale
1m :1cm
1:100
X= √(R 2 -Y 2 )-
Offset, Y √(R 2 -(W/2) 2 )
0 3.515
2 3.347
4 2.828
6 1.907
8 0.459
W/2 = 8.845m W/2 = 8.845m
8.485 0.000
T1O 8.845m
O 8m
O 6m
O 4m
O 2m
O 0m
O 2m
O 4m
O 6m
O 8m
O 8.845m
T 2
2.828m
1.907m
2.828m
3.515m
3.515m
W = 16.971 m
3.515m
2.828m
1.907m
2 m 2 m 2 m 2 m 0.459 m
2.828m

Given data of curve ranging was as follows:-
Radius = 650 m
Deflection angle = 17 0 58’50”
Offset = 20m
Chainage 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.

Radius, R
Deflection Given angle ,
θ
Offset
Chainage
intersection
point, I
formula
1718.9 x C
δ
1
=
(deg ree )
60R
1718.9 x C
δ
1
=
(minute
)
R
650m
17º 58’ 50”.
20m
4100m
Stn
.
Chainag
e
T1
Draw the table form for
deflection angle method
Chord
length
I
θ = 17º 58’ 50”.
Deflection
angle,δ
(0 ‘ “)
T2
R = 650
m
Setting out
angle, δ
(0 ‘ “)

PROCEDUR
Tangent length = R tan θ/2
= 650 tan (17º 58’ 50”/2) = 102.837m
1718.9 x 2.837
Chainage T 1
= chainage I – tangent length δ =
E
= 4100.00 - 102.837 = 3997.163m 60x650
Arc length
= R x θ x 2π
1718.9 x 20.000
360
δ =
= 650 x 17 o 58’50” x 2π = 188.292m 60x650
360
Chainage T 2
= chainage T 1
+ arc length
= 3997.163 + 188.292 = 4185.455m
1718.9 x 5.455
=
60x650
Stn. Chainage Chord length, C Deflection angle,δ Setting out angle, δ
T1 3997.163 0 0 0 0’ 0” 0 0 0’ 0”
δ1 4000 2.837 0 0 19’ 30” 0 0 19’ 30”
δ2 4020 20.000 0 0 52’ 53” 1 0 12’ 24”
δ3 4040 20.000 0 0 52’ 53” 2 0 5’ 17”
δ4 4060 20.000 0 0 52’ 53” 2 0 58’ 10”
δ5 4080 20.000 0 0 52’ 53” 3 0 51’ 4”
δ6 4100 20.000 0 0 52’ 53” 4 0 43’ 57”
δ7 4120 20.000 0 0 52’ 53” 5 0 36’ 50”
δ8 4140 20.000 0 0 52’ 53” 6 0 29’ 44”
δ9 4160 20.000 0 0 52’ 53” 7 0 22’ 37”
δ10 4180 20.000 0 0 52’ 53” 8 0 15’ 31”
T2 4185.455 5.455 0 0 37’ 30” 8 0 53’ 1”
δ
Σ = 188.292 Σ = 8 0 53’ 1” θ / 2 = 17 0 58’50” / 2 = 8 0 53’ 1”

Radius, R 24.7m
Deflection angle , θ 60 0
Offset
5m
Chainage intersection point, I 20 m
Tangent length = 14.261m
Chainage T 1
= 5.739m
Arc length =
25.866m
Chainage T 2
=
31.605m
Stn. Chainage Chord length, C Deflection angle,δ Setting out angle, δ
T1 5.739 0 0 0 0’ 0” 0 0 0’ 0”
δ1 10 4.261 4 0 56’ 32” 4 0 56’ 32”
δ2 15 5.000 5 0 47’ 57” 10 0 44’ 29”
δ3 20 5.000 5 0 47’ 57” 16 0 32’ 26”
δ4 25 5.000 5 0 47’ 57” 22 0 20’ 23”
δ5 30 5.000 5 0 47’ 57” 28 0 8’ 20”
T2 31.605 1.605 1 0 51’ 42” 30 0 0’ 2”
Σ = 25.866 Σ = 30 0 00’ 2” θ / 2 = 60 0 / 2 = 30 0

Scale
1m :1cm
1:100
I
14.261 m = 14.261cm
T1
T2

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

Given data of curve ranging was as follows:-
Radius = 600 m
Deflection angle = 18 0 24’
Chainage I = 2140m
Based on data-data given above,
•Provide a table of setting out by two theodolite without measuring tape.

PROCEDUR
Tangent length = R tan θ/2
= 600 tan 18°24′/2= 97.20m
1718.9 x 17.179
Chainage T 1
= chainage I – tangent length δ =
E
= 2140.00 - 97.20 = 2042.80m 60x600
Arc length
= R x θ x π
1718.9 x 20.000
360
δ =
= 600 x 18 o 24’ x π = 192.684m 60x600
360
Chainage T 2
= chainage T 1
+ arc length
= 2042.80 + 192.68 = 2235.48m
Stn. Chainage Chord length, C Deflection angle,δ
(0 ‘ “), T1
T1 2042.821 0 0 0 0’ 0”
δ1 2060 17.179 0 0 49’ 12”
δ2 2080 20.000 0 0 57’ 18”
δ3 2100 20.000 0 0 57’ 18”
δ4 2120 20.000 0 0 57’ 18”
δ5 2140 20.000 0 0 57’ 18”
δ6 2160 20.000 0 0 57’ 18”
δ7 2180 20.000 0 0 57’ 18”
δ8 2200 20.000 0 0 57’ 18”
δ9 2220 20.000 0 0 57’ 18”
T2 2235.506 15.506 0 0 44’ 25”
δ
Σ = 192.684 Σ = 9 0 12’ 1”
1718.9 x 15.506
=
60x600
Deflection angle,δ
(0 ‘ “), T2

Stn. Chainage Chord length, C Deflection angle,δ
(0 ‘ “), T1
360° - θ + δ 1
2
Example δ1
= 360° - θ + δ 1
2
= 360° - 18 0 24’ + 0 0 49’ 12”
2
= 351° 37’ 20”
Deflection angle,δ
(0 ‘ “), T2
T1 2042.821 0 0 0 0’ 0” 350° 48’ 00”
δ1 2060 17.179 0 0 49’ 12” 351° 37’ 20”
δ2 2080 20.000 0 0 57’ 18” 352° 34’ 40”
δ3 2100 20.000 0 0 57’ 18” 353° 32’ 00”
δ4 2120 20.000 0 0 57’ 18” 354° 29’ 20”
δ5 2140 20.000 0 0 57’ 18” 355° 26’ 20”
δ6 2160 20.000 0 0 57’ 18” 356° 23’ 40”
δ7 2180 20.000 0 0 57’ 18” 357° 21’ 00”
δ8 2200 20.000 0 0 57’ 18” 358° 18’ 20”
δ9 2220 20.000 0 0 57’ 18” 359° 15’ 40”
T2 2235.506 15.506 0 0 44’ 25” 360° 00’ 00”
Σ = 192.684 Σ = 9 0 12’ 1”