Sub: SurveyingTopic : Theodolite
MODULE- 97C
Theodolite traversing
________________________________________________________
1.
The horizontal distance between two stations P and Q is 100m. the
vertical angles from P and Q to the top of a vertical tower at T are
and
above horizontally respectively. The vertical angles from
P and Q to the base of tower are
and
below horizontal
respectively. Station P, Q and the tower are in the same vertical
plane with P and Q being on the same side of T. neglecting earth’s
curvature and atmospheric refraction, the height in m of the tower
is
a) 6.972
b) 12.387
c)
12.54
d) 128.745
Ans:
(
)(
d=128.746 m
∴h=128.746(tan5+tan0.5)
=12.387 m
2.
)
(
)
A theodolite is set up at station A and a 3m long staff is held
vertically at station B. the depression angle reading at 2.5m
marking on the staff is
. The horizontal distance between A
and B is 2200m. Height of instrument at station is 1.1m and R.L of
A is 880.88m. Apply the curvature and refraction correction and
determine the R.L of B in (m) .
a) 641.454 b) 639.367
c) 640.170 d) 642.980
Ans:
R.L of B= 880.88+1.1-237.7-2.5=641.78
Correction for curvature =
JH ACADEMY
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Sub: SurveyingTopic : Theodolite
( )
Refraction= (-)
0.0673
3.
The observation from a closed loop transverse around the obstacle
are
Segment
Observation length
From
Azimuth
(clockwise
Station
from magnetic north)
PQ
P
MISSING
33.
QR
Q
300.000
86.384
RS
R
354.524
169.381
ST
S
450.000
243.900
TP
T
268.000
317.500
What is the value of the missing measurement (rounded off to the
nearest 10mm)?
a)
396.86m
b) 396.79m
c) 396.05m
d) 396.94m
Ans:
18.917
-348.45
-197.97
197.59
0.8131x - 329.917
299.403
65.325
-404.113
-181.058
0.556x – 220.443
x = 396.79 m
4.
The length and bearings of a closed transverse PQRS are given
below
Line
length (m)
bearing (WCB)
PQ
200
QR
1000
RS
907
SP
?
?
The missing length and bearing, respectively of the line SP are
a) 207m and
b) 707m and
c) 707m and
d) 907m and
JH ACADEMY
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Sub: SurveyingTopic : Theodolite
Ans:
Latitude
Departure
200
707.1
- 907
0
0
707.1
0
707.1
∴
=0
= -707.1
∴ θ = 270
= 707.1
5.
The focal length of the object glass of a tacheometer is 200mm, the
distance between the vertical axis of the tachometer and the optical
center of the object glass is 100mm and the spacing between the
upper and lower line of the diaphragm axis is 4mm. with the line of
collimation perfectly horizontal, the staff intercepts 1m (top), 2m
(middle) and 3m(bottom). The horizontal distance (m) between the
staff and instruments station is
a) 100.3 b) 103.0
c) 150.0
d) 153.0
Ans:
=
=
+
–1
=
or
=
=
∴ = 200(1 + 500)
= 501 ×
= 100.2 m
Distance between staff and instrument is 100.2 +0.1 = 100.3m
6.
The following table gives data of consecutive co-ordinates in
respect of a closed theodolite transverse PQRSP
Station
Northing Southing Easting Westing
P
400.75
Q
100.25
300.5
199.25
R
199.0
S
300.0
299.75
200.5
The magnitude and direction of error of closure in whole circle
bearing are --------a)
2.82 m and 315
b) 2.84 m and 305
c)
2.80 m and 325
d) 2.80 m and 300
JH ACADEMY
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Sub: SurveyingTopic : Theodolite
Ans:
∑
= 501,
√
7.
∑ = 499,
∑ =499,
∑
=501
∴ 2.82 m and
and
The bearings to two inaccessible stations A and B taken from
station were
and
respectively. The co-ordinates
of A and B were as follows
Station
easting
northing
A
300
200
B
400
150
Calculate the independent co-ordinates of C.
a) 362, 222.59 b) 360, 222.79
c)
364, 223.59
d) 361, 221.59
Ans:
p
+q
= 100
p
-q
= -50
0.9396 p + 0.4628 q = 100
0.342 p – 0.886 q = -50
P = 66.07
∴ x = 300 + 66.07
= 362
∴ y = 200 + 66.07
= 222.59
“ C ” co-ordinates (362, 222.59)
8.
A tachometer fixed with stadia wires 4mm apart has its object glass
(f=200mm) fixed at a distance of 250mm from the trunnion axis.
The tachometric distance equation is
a) D=100S+0.45
b) D=50S+0.45
c) 100S+0.25
d) 100S+0.35
Ans:
=
+
=
=
(
or
= 1+
=
-1
–1
)
+ d = D = . S +(f +d)
∴
9.
. S + 0.20 +0.25 = 50S + 0.45
Two tangents spaced 6.0m apart were fixed on a sub tense bar and
the vertical angles measured on the two upper and lower targets
were
and
respectively. If the lower target was
at an elevation of 249.2m what was the height of instrument
a) 240.45 b) 239.45
c) 241.45
JH ACADEMY
d) 242.45
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Sub: SurveyingTopic : Theodolite
Ans:
=
=
∴ 0.0435 =
0.0258 =
∴
∴6
0.0258 + 0.0258V = 0.0435V
V = 8.746m
d = 339 m
Level of collimation = 249.2 – 8.746
= 240.454
10. The constants for an instrument are 1000 and 0.5.calculate the
distance from the instrument to the staff when the micrometer
readings are 3.246 and 5.246. the staff intercept is 2.0m and
vertical angle measured is
, the staff being vertical
a) 189.95
b) 188.95
c) 190
d) 200
Ans:
D=
=
JH ACADEMY
.S
+C.
.2×
+ 0.5 ×
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