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
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 =
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 ‘ “)
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.