Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Vertical section
Sheet No: 1
(1) The following leveling observations were obtained along the center line of a proposed line
route:Staff distance
R.L.
Remarks
0
178.69
Point A
20
178.35
40
177.89
60
178.29
80
178.52
100
178.79
120
178.59
140
179.80
160
180.16
Point B
If the pipe line is to start at a reduced level 177.00 m at A and rise at a gradient of 1: 100
towards B.
(a) Draw a vertical section along the proposed route with suitable horizontal and vertical
scale, taking 173.00 at the datum of heights.
(b) From the graph or by calculation determine the cut and fill at each point
(2) The under noted staff readings were taken successively with a level along an underground
roadway.
Distance from A
St. reading
Remarks
L
C
R
1.759
0
BS to A
0.652
20
IS
1.541
40
5
IS
1.666
40
10
IS
0.985
3
40
IS
1.245
15
40
IS
0.091
40
FS
1.689
BS
0.910
80
FS
0.680
BS
0.917
120
IS
0.650
160
FS
0.671
BS
1.540
200
IS
1.955
240
FS
Using the H.I method, Calculate:
(a) The reduced level of each staff station relative to the level of A, which is 120 m,, and
check the results.
(b) Draw a profile section between points A and B with horizontal scale 1:1000 and a vertical
scale of 1:10.
(c) Compute the cut and fill if the constructed roadway is to have a constant elevation
between A and B of 122 m.
(d) Draw a cross-section at interval 40 m.
1 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Cross-section
Sheet No: 2
(1) An embankment of width 10 m and side slopes 1.5: 1 is required to be made on a
ground which is level in a direction transverse to the centre line. The central heights at
40m intervals are as follows:
0.90 - 1.25 - 2.15 - 2.5 - 1.85 - 1.35 and 0.85
Calculate the volume of earth work according to
(a) The trapezoidal formula
(b) The primordial formula.
(2) Railway embankment of formation with of 8m and side slop 2:1 is to be
constructed .the ground level along the centre line is as follows:
Chainage- 0
50
100
150
200
250
GL (m) - 115.75
116.35 116.80
116.50 115.50
115.25
The embankment has a rising gradient of 1 in 100 and the formation level at zero
chainage is 115.00 Assuming the ground is level across the centre line compute the
volume of earth work.
(3) The ground level along the centre line of a road is given below: Chainage- 0
50
100
150
200 250
300
GL (m) - 117.50 116.25 115.95 116.65 117.20 117.85 115.75
It is proposed that the formation level of RL 115.00 should be kept constant of starting
from the chainage zero the formation width of the road is 8 m and the side slope 1:1
the ground is level transverse to the centre line.
(4) The formation width of a certain cutting is 8m and the side slope is 1:1 the surface
of the ground has a uniform slope 1 in 10 if the depths of the cutting at centers of three
sections 40m apart are 2, 3 and 4m respectively find the volume of earth work .
2 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Theodolite
Sheet No: 3
1- Discuss the various parts of theodolite and illustrate your answer with
sketch.
2- Define the following terms:a- Centering
b- Face left
c- Face right
d- Line of collimation
e- Geometrical axis
f- Optical axis
g- Line of sight
3- Explain briefly the temporary adjustment of a theodolite.
4- Name the fundamental axes of a theodolite. State the relationship that
should exist between them when the instrument is in adjustment.
5- Describe the following:a- Plate bubble adjustment.
b- Collimation adjustment.
c- Trunnion axis adjustment.
d- Diaphragm adjustment.
e- Index error
6- In the horizontal angle ABC = 115°20` 15``, AB is 700 m and BC is
1000 m. IF the error in centering the target at B is  5mm, what will be the
resultant error in the measured angle ? 
3 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Horizontal and vertical angles
Sheet No: 4
Several angles have been measured at survey station A. The following data refer to
the angles: Instrument
Target St.
Horizontal Circle Reading
station
FL
FR
A
B
00° 24`40``
180°24`40``
C
65 36 00
245 36 20
D
132 10 20
312 10 40
E
247 08 00
67 08 20
Calculate: a- The mean observed angles BAC, CAD, and DAE.
b- The bearing of AB, AC and AD if the bearing of AE = 2400 00\ 00\\
2- A theodolite was used to observe the top of a spire (S) from two stations(A) and (B)
whose distance was 101.8 m , and the observations at the stations were as follows :Station
To Face
H.C.R.
V.C.R.
S
L
000°
00`
40`` 068°
36`
34``
R
180°
00`
36`` 291°
23`
22``
B
L
058°
12`
58`` 089°
13`
25``
A
R
238°
12`
54`` 270°
46`
23``
S
L
000°
00`
42`` 068°
38`
38``
R
180°
00`
38`` 291°
21`
20``
S
L
000°
00`
40`` 068°
49`
17``
R
180°
00`
34`` 291°
10`
35``
B
A
L
064°
17`
24`` 092°
24`
11``
R
244°
17`
20`` 267°
35`
55``
S
L
000°
00`
43`` 068°
50`
16``
R
180°
00`
39`` 291°
09`
38``
Compute the mean height of the spire above the ground if the height of the instrument
was 1.46 m above each of the two stations. The two stations have the same elevations,
and find the index error.
3- The angle is measured by repetition transit. The mean back sight reading is 359°
59`45``.After the first repetition the reading is72°29`05``. After the twelfth repetition, the
reading is 149° 47` 12``. Compute the value of the angle.
4- A theodolite was set over station (X) to measure directions to A, B, and C. The
following observations were determined:
Station
To
Face
V.C.R
occupied
A
L
85O 30' 10''
R
274 29 40
B
L
95 30 10
R
264 29 44
C
L
85 30 15
R
274 29 35
Compute:
a- The corrected vertical angle.
b- The Index error.
4 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Omitted observation
Sheet No: 5
1- The following table refers to a closed traverse ABCDEA .The bearing
of the line DE and length of line CD could not be observed.
Calculate the omitted measurements.
Line
AB
BC
CD
DE
EA
Length
800
700
???
790
500
Bearing
N 35 ° 00` E
S 85 00 E
S 03 00 E
?? ?? ??
N 60 00 W
2- Lengths of all the lines, reduced bearings, eastings and northings of
AB, DE and EA lines of a closed traverse ABCDEA are given as
follows. Determine the bearings of BC and CD.
Line
AB
BC
CD
DE
EA
Length
365.0
525.5
622.5
470.0
287.5
R.B.
S 60° 00` E
---S 49 22 E
----S 02 24 W
3- Using the data of a closed traverse given below, calculate the lengths
of the lines BC and CD.
Line Length
W.C.B.
AB
344
014° 31`
BC
L1
319 42
CD
L2
347 15
DE
300
05 16
EA
1958
168 12
5 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Link Traverse
Sheet No: 6
1- Give the measurements of the traverse 1234, as shown in the Fig. (1), and the data,
is given in the following table:
North
East
Station
9491.70
4206.42
A
9342.21
4284.18
B
8477.12
5606.21
C
8185.03
5577.30
D
Compute the balanced co-ordinates for the traverse stations (1-2-3-4).
2- The following is an extract form the traverse calculations between two fixed points
P and Q.
Line Length ∆E
∆N
PA
86.19
-1.84
+86.17
AB
69.08
-61.52
-31.41
BC
145.4
-21.50
+151.42
CQ
82.24
-79.85
-19.68
The known correct differences in easting and northings between stations P and Q are
respectively –164.02 and 186.15. Calculate the correct differences of coordinates of
the traverse using Transit rule.
3- The following data were obtained in a link traverse. Calculate the final adjusted
partial co-ordinate using transit method.
Side
Station Observed angles
Length
T2
131° 25` 20``
T2A
147.65
A
138 37 00
AB
139.10
B
147 43 20
BC
111.57
C
261 21 40
CT3
121.33
T3
259 33 00
The co-ordinates of T2 and T3 in meters are
T2 = 403.00 E; 351 N
T3 = 847.86 E; 335N
The known bearings are
T2T1 = 345° 00` 00``
T3T4 = 203° 42` 00``
Fig. (1)
6 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Tacheometry (1)
Sheet No: 7
1- A line was leveled tacheometrically with a tacheometer fitted with an analectic
lens, the value of the constant being 100. The following observations were
made, the staff having been held normal to the line of sight.
Instrument
V.A.
H.I
Staff at
Staff reading
Remarks
Station
1.50
- 2° 54` 1.02 , 1.720 , 2.42
A
B.M
R.L. = 180
1.50
+
3
36
1.22
,
1.825
,
2.43
A
B
1.42
+ 4 06
0.78 , 1.610 , 2.44
B
C
Compute the distances AB, BC, and A to the B.M. Also find the
elevations of A, B, and C.
2- Two sets of tacheometric reading were taken from an instrument station A, the
reduced level of which was 100.06 m, to a staff station B.
a- Instrument P- multiplying constant 100, additive constant 0.25 m.
b- Instrument Q- multiplying constant 90, additive constant 0.25 m.
Instrument At
To
H.I.
VI.A.
Stadia readings
1.50
26°
14`
22``
0.755
,1.005 ,1.255
P
A
B
1.45
25 17 32
??? ??? ???
Q
A
B
What should be the stadia readings with instrument Q?
3- Two sets of tachometric readings were taken from an instrument station A (R.L.
100.00), to a staff station B:
Instrument
P
Q
Multiplying Constant 100
95
Additive constant
30 cm
45 cm
Height of instrument 1.40 m
1.45 m
Staff held
vertical
normal to line of sight
Instrument
At To Vertical angle Stadia readings
P
A
B
5° 44`
1.090/1.440/1.795
Q
A
B
5° 44`
???
Determine:a- The distance between instrument station and staff station.
b- The R.L. of staff station B.
c- The stadia readings with instrument Q.
4- A tacheometer with a multiplying constant of 100 and no additive constant is set
up over station X and sighted on a level staff held vertically on a bench mark B , and
then on two points Y and Z . The height of B above datum is 278.36 m. The results
are in the following table:
Instrument Station
Vertical
Horizontal
Stadia hair readings
station
sighted Upper Center Lower
angle
circle reading
2.310
1.887
1.464
+ 2° 15` 10``
------X
B
2.466
1.600
0.732
+ 3 36 00
28° 12` 15``
X
Y
2.522
1.455
0.388
2
42
30
93 51 55
X
Z
Calculate: (a) The horizontal distance YZ.
(b) The height of points Y and Z.
7 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Tacheometry (2)
Sheet No: 8
1- In order to determine the constants of a tacheometer, distances 200 and 400 m were
accurately measured and readings on a stadia rod on the upper and lower wire were
taken as follows:
At 200 m
2.00
4.00
At 400 m
0.50
4.50
Determine the values of the constants and find the distance when the readings of
the other stadia wires were 1.5 and 4.5 m. The line of sight being horizontal in all
cases.
2- A vertical staff is observed with a horizontal external focusing
telescope at a
distance of 112.489 m, measurements of the telescope
are recorded as:
Objective to diaphragm
230 mm
Objective to vertical axis
150 mm
If the readings taken to the staff were 1.073, 1.626 and 2.185, Calculate:
a- The distance apart of the stadia lines.
b- The multiplying constant.
c- The additive constant.
3- A tacheometer was set up at a station A and the following readings were obtained
on a vertically held staff .
Station Staff station V. angle
Stadia readings
Remarks
A
-2
°
18``
3.225,
3.550,
3.875
R.L.
of B.M. is
B.M
8° 36``
1.650, 2.515, 3.380
200 m
B
Calculate the horizontal distance from A and B and the R.L. of B, if the constants of
the instrument were 100 and 0.425 m.
4- The following observations were taken from two traverse stations by means of a
tacheometer fitted with an analectic lens. The constant of the instrument is 100.
Instrument
Staff
H.I
Vertical
Bearing Staff readings
station
station
angle
A
C
1.38
+10° 12`
226°30 0.765 /1.595 /2.425
B
D
1.42
- 12 30
84 45 0.820 /1.840 /2.860
Co-ordinates of station A = 12.30 N, 186.8W
Co-ordinates of station B = 102.8 N, 96.4W
Compute the length and gradient of the line CD, if B is 6.50m higher than A.
5- Two observations are taken upon a vertical staff by means of theodolite, of which
the R.L. of the horizontal axis is 250.0 m. In the first observation , the line of sight is
direct to give a staff reading of 1.00 m and an angle of elevation is 4° 58` . In the
second observation, the staff reading is 3.50 m and the angle of elevation is 5° 44`.
Compute the R.L., of staff station and its horizontal distance from the instrument.
8 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Rankin's method
Sheet No: 9
1- Draw a neat sketch of a simple circular curve and show the following elements
there on. Give their relationships.
a- back tangent
b- forward tangent
c- Point of intersection
d- point of tangent
e- Angle of intersection
f- angle of deflection
g- Long chord
h- central angle
i- Point of curvature
j- radius of curve
2- If the radius of a circular curve is 400 m and angle of intersection is 120°. Find
the various elements of a simple circular curve, taking the chainage of the
point of intersection as 1220.5 m.
3- Two straight lines AB and BC intersect at chainage 2067 m, the intersection
angle being 140°. It is desired to connect these two straight lines by a simple
circular curve of 5°. Calculate the radius of the curve and tabulate the
necessary data to set out the curve with a peg interval of 30 m.
4- Tabulate the necessary data to set out a right – handed simple circular curve of
250 m radius connecting two straight lines having a point of intersection at a
chainage of 3450 m using Rankin’s method. The intersection angle between
the two straight lines is 50° .Take the chord interval as 20 m.
5- In setting out a circular road curve it was found necessary that the curve
should pass through a point at a distance of 30 m from the intersection point
and equidistant from the tangents . The chainage of intersection point was
1728 m and the intersection angle 15°. Find out the necessary data to set out
the curve with a peg interval of 30 m.
6- 6- Two straights intersecting at a point B have the following bearings, BA
270°, BC 110°. They are to be joined by a circular curve which passes through
a point D which is 150 m from B and the bearing of BD is 260°. Find the
required radius, tangent lengths, length of curve and setting out the first four
points using 30 m standard chord.
9 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: deflection distance method
Sheet No: 10
1- Describe briefly the method of setting out a simple circular curve with help of two
tapes.
2- Tabulate the necessary data to set out a right – handed simple curve of 250 m radius
connecting two straights having a point of intersection at a chainage of 3450 m
by the method of offset from the chords produced. Take the chord interval as
20 m. The deflection angle between the two straights is 50°.
3- Two straights AB and BC intersect at a chainage 3813.6 m. The angle of
intersection is 140°. It's required to set out a 5° simple curve to connect the two
straights. Calculate all the data necessary to set out the curve by the method of
offsets from the chords produced with a peg interval of 30 m.
4- A curve of radius 500 m is to be set out to join two straights BI and IC which
deflect through 120°. Point I is inaccessible and the following data was
measured:
BC = 525.2 m, angle IBC = 65° 30` 00``.
Assuming the chainage of B is 483.65 m, calculate sufficient data to set out
the curve by deflection angles from the tangent point T1 by standard chords of
40 m.
10 / 11
Higher Technological Institute
Civil Engineering Department
6th of October
Course Title: Surveying (ii)
Topic: Vertical curve
Sheet No: 11
1- What are the properties of the parabolic vertical curve?
2- What is the difference between summit and valley carves and why are they needed?
3-What information and formulae are necessary in order to calculate the setting out data
of a vertical curve?
4- An existing road of rising gradient 2.5% meets a falling gradient 2%, at chainage
364.370 m and level 50.36 m. The gradient is to be connected by a simple
parabolic curve 100 m long. Calculate:(a) The rate of change of gradient.
(b) The chainage and levels of the tangent points.
(b) The levels of the points on the curve using the standard chord of 20 m.
5- As part of dual highway reconstruction scheme, the following line of levels was
taken at given points on the existing surface.
Point Reduced level Chainage (m)
31.891
632.46
A
33.263
678.18
B
32.702
777.24
C
31.635
830.58
D
If the curve is designed to give a rate of change of gradient of -2.5 x 10-4, Calculate:(a) The length of the curve L.
(b) The chainage and level of intersection point (I).
(c) The chainage and levels of the tangent points T1 and T2.
(d) The level of the points on the curve (based on standard chords of 20 m).
6- Two straights T1I and IT2 falling to the right at 1% and 5% , respectively are to be
connected by a parabolic vertical curve 200 m long . Given that the chainage and
reduced level of I are 3627.00 m and 84.64 m, respectively. Design the vertical
curve using the standard chord of 20 m.
7- A down grade of 1.5 % is followed by an upgrade of 1:40. Take RL. Of the point of
intersection is 60.00 and its chainage is 360 m. A vertical curve 120 m long is to
be introduced to connect the two grades. The pegs are to be fixed at 15 m
intervals. Calculate the elevation of the points on the curve and work out the
staff readings required if the pegs are to be driven with their tops at the
formation of curve. It is given that the height of collimation is 62.180 m.
11 / 11