Basic concepts of surveying

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 Learning outcomes (objectives)
• Measure distance with linear measuring
instruments
• Set up and use of
 Leveling
 Theodolite
 Total station and
 GPS device
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 Introduction
• Land is all around us. We walk on it,
build houses and commercial buildings
on it, drive on it, fence it, dig trenches in
it, and farm it. Life as we know it would
not exist without land. Its presence
permeates our lives to the point that we
seldom think about it. Yet in order to plan
or analyze the use of land, measurements
must be made.
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 Cont…
• Many of our uses of the land require us to
measure, mark, or locate points on, above or
below the surface. We often do this without
thinking about the principles we are using.
Property is located and marked before a fence is
built.
• A carpenter carefully marks the corners of a
building before starting construction.
• Engineers and planners may spend months
deciding on the location of a road and marking it
out.
• The slope and other features of an area must be
measured before a pond is built or drainage way
constructed.
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 Cont…
• The property corners of a parcel of land are located and marked
during the transfer of ownership. The principles of surveying are
used in all of the examples mentioned above.
• The complexity of surveying can range from taking a few
minutes and two sticks to lay out a 90-degree corner, to spending
several days with thousands of dollars worth of equipment
establishing a road or power line right of way to establishing
survey control monuments, the most complex survey.
• This training will define surveying and some of the essential
terms used in surveying.

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 Definition
• Surveying may be simply defined as the art
of making measurements.
Distances
angles and
Elevations
• surveying includes the computation of
Areas
Volumes and
other quantities, as well as the preparation of
necessary maps and diagrams.

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 Cont…
• The survey activities are not limited to the
surface of the earth but extend to the sea and
deep underground, as well as extraterrestrial
space.
• More precisely, Surveying can be defined as
the art and Science of determining the relative
positions of various points on, above or below
the surface of the earth.

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 Cont…
• relative positions :-determining the relative
positions of points in reference to another
point or reference point/station

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 The main objectives of any surveying
activities
• To measure the horizontal distances between the
points.
• To measure the vertical elevations between the
points.
• To determine the relative direction between the
points/ lines by measuring the horizontal angles with
reference to any arbitrary (approximate/ unscientific)
direction
• To determine the absolute direction by measuring the
horizontal angles with reference to any fixed
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Importance of Surveying to Civil Engineers
– To establish the national and state boundaries.
– To chart coastlines, navigable streams and lakes;
– To fix the control points.
– To plot hydrographic and oceanographic charts and maps
– To prepare topographic map of land surface of the earth.
The ultimate results of surveying include;
• To obtain data from field/site
• To generate map or plan of the surveyed area
• To compute and analyze the field data for setting out
the operation of actual engineering works
• To establish field parameters in the site for further
usage
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Primary Division of Surveying

i. Plane surveying
ii. Geodetic Surveying

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 Cont…
• Plane surveying- is the type of surveying in which the
curvature of the earth is neglected and it is assumed to be
a flat surface. All horizontal distances and horizontal
angles are assumed to be projected on a horizontal plane.
A horizontal plane at a point is the plane, which is
perpendicular to the vertical line at that point.
• Geodetic Surveying: It is the type of surveying in which
the curvature of the earth is taken in to consideration and
a very high standard of accuracy is maintained. The main
objective of geodetic surveying is to determine the precise
location of a system of widely spaced points on the
surface of the earth.
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 Classifications of Surveying based on Application
i. Based on the nature of field
ii. Based on the objective / purpose
iii. Based on methods used
iv. Based on instruments used
1. Classification based upon nature of field
• Land survey
• Marine Survey
• Astronomical Survey

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2. Classification based on objective / purpose:
1. Property survey
2. Topographic Survey
3. Construction Survey
4. Control Survey
5. Route Survey
6. City surveys
7. Mine surveys
8. Hydrographic surveys
9. Engineering surveys
10.Astronomic surveys
11.Satellite surveys
12.Geological surveys
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 Cont…
• A property survey is performed in order to establish the
positions of boundary lines and property corners. It is also
referred to as a land survey or a boundary survey.
• City surveys are conducted within the limits of a city for
urban planning. These are required for the purpose of layout
of streets, buildings, sewers, pips, etc.
• Engineering surveys are conducted to collect data for the
designing and planning of engineering works such as
building, roads, bridges, dams, reservoirs, sewers and water
supply lines.
• Astronomic surveys are conducted for the determination of
latitudes, Azimuths, local time etc. for various places on the
earth by observing heavenly bodies (the sun or stars).
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 3. Based on methods used
• Triangulation surveying:
• Traverse surveying
• Property surveying: A property survey is performed in order
to establish the positions of boundary lines and property
corners.
• Topographic Surveying: A topographic survey is performed
in order to determine the relative positions (horizontal and
vertical) of existing natural and constructed features on a tract
of land.
• Construction Surveying: A construction survey, also called
a layout or location survey is performed in order to mark the
position of new points on the ground.

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 4. Based on instruments used
• Leveling
• Theodolithe
• Totalstation
• GPS
• .
• .

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Classification of Surveying based on
Instruments used

i. Chain/ tape Surveying

ii. Compass surveying
iii. Leveling
iv. Theodolite surveys
v. Tachometric Survey
vi. Photogram metric survey
vii. EDM Surveys:
viii.Plane table surveys:

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 Units of measurement

There are two main systems of measurements:

1) M.K.S. metric system (SI=System
International)
2) F.P.S. or British system

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 The SI unit for area is the derived units
square meter (m2)

• 1 hectare = 104m2 =
= 100 m * 100m = 1 ha
• 1 square kilometer = 106m2 =
= 1000m * 1000m = 1km2 = 100ha

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 The SI unit for volume is the derived
unit cubic meter (m3)
• 1000 cu millimeters = 1 cubic centimeter
• 1000 cu centimeters = 1 cubic decimeter
• 1000 cu decimeters = 1 cubic meter

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 The SI units for plane angles
• There are three systems in use for angular unit,
namely
• Sexa-gesimal graduation (degrees)
• centesimal graduation and (gon)
• radian (the full circle equals 2 rad.)

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 Conversion of units
a) Conversion of length c) Conversion of volume
• 1 inch = 2.54 cm 1 cu in = 16.387 cu cm
• 1 foot = 0.3048m 1 cu ft = 0.0283 c um
• 1 mile = 1.6093 km
d) Conversion of angles
b) Conversion of area 1 gon = 9/10 deg
 1 sq in = 6.4516 sq cm 1 deg = 10/9 gon
 1sq ft = 0.0929 sq m
 1 sq mile = 2.59 sq km

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 Activity 1
Convert one unit to the other units
1. from gon to degree: 48.0488 gon
2. from degree to gon: 43.2439
3. from degree decimal value to degree,
minutes, seconds: 43.2439
4. degree, minutes, seconds to degree decimal
value: 43 14’ 38”

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 Solution
1. 48.0488 gon * 9/10 = 43.2439
2. 43.2439 * 10/9 = 48.0488 gon
3. 43 + 0.2439
 0.2439 * 60 = 14.6340’
 0.6340’* 60 = 38”
 result: 43 14’ 38”
4. 43 + 14’/60 + 38”/ 3600 = 43.2439

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 Scales of a map
• Scale of a map is the ratio of the distance
marked on the map to the corresponding
distance on the ground.
• Scales of a map are generally classified as
large, medium and small.
 Large scale 1:1000 or less
 Medium scale 1:1000 _ 1:10,000
 Small scale 1:10,000 or more

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 Representation of scale

i. By statement (engineer’s scale)

ii. By representative fraction (R.F)/
Ratio
iii. By Graphical scale

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• Statement scale:-Specially for people who are
not familiar with mathematics.
• R.F scale:-It is the most universal and logical
way of expressing scales
• Graphical scale:-It is the easiest of all scales to
use because no calculations are involved when
you want to find the real distance b/n two
points, you don’t even need a ruler to find the
distance.

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Graphical scale

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 Activity 1
a, Change 1:250,000 to Statement Scale
B, Change 1:5000,000 1cm to graphic scale
C, Change 1cm to 2km to R.F scale
D, The ground distance is 10km for 2 cm map
distance

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 Solution
A, 1:25000 ,1cm to 250000 cm , 250000cm = 2.5 km
:. 1cm to 2.5 km.
B, 1:500,000 500,000 cm
1cm to 5km.
C, 1cm to 2km 1cm = 200,000cm
1:200000 or 1/200000
D, - 10km = 1,000,000 cm
The division of the line equals 2cm
The scale statement is 2cm to 1,000,000cm
2/1000000
Writing in ratio, = 1:500,000
= 1:500,000
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Basic Measurements in Surveying

i. Horizontal distance
ii. Vertical Distance
iii.Slope distance
iv.Horizontal Angles and
v. Vertical Angles

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 Measurement of measuring Distances:
Horizontal Distances: Vertical Distances:
• Direct methods; • Barometric leveling
• Plastic tube leveling
• Optical methods; and
• Trigonometric leveling
• Electronic method. • Differential leveling
• GPS

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 Linear Measurements
Introduction
Every surveyor has to measure the horizontal
distance between two points on the surface of
the earth. Measurement of horizontal distance
or making linear measurements is required in
chain surveying, traverse surveying, and other
types of surveying.

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 Cont…
• In surveying the distance between two points
means a horizontal distance.
• When slope distances are measured in the
field, these are always reduced to the
equivalent horizontal distances for preparation
of map.
• There are, in general, three methods of making
linear measurements:
1. Direct methods.
2. Optical methods/ Computational/ Indirect Method
3. E.D.M. methodsprepared By Afework Legesse
 Cont…
• In the direct methods, the distance is actually
measured in the field using a tape.
• In optical methods, the distance is not actually
measured the field. It is compute indirectly.
• Electromagnetic Distance measuring (E.D.M)
instruments have been developed quite recently.
These are basically of two types:
• Electro optical instruments, which use light waves
for measurement of distance.
• Microwave instruments, which use radio waves
for measurement of distance.
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 Approximate method
• commonly used in reconnaissance surveying for the
measurement of horizontal distances
1. Pacing :-A person can determine the distance
walked by counting the number of paces made.
The distance can be obtained by multiplying the
number of paces by the average length of the
paces
2. Measuring wheel. A measuring wheel consists of a
wheel mounted on a lower end of rod about 1 m
long through a fork. The upper end of the rod has
a handle. The wheel is pushed along the ground.
The distance traversed is recorded on the dial
attached the wheel.
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 Cont…
3. Speedometer. All automobiles have a speedometer
to indicate the speed and the distance traveled. If
the ground is smooth, the speedometer can be used to
measure the distance.
4. Passometer. It registers the number of paces the
mechanism operator automatically due to the motion
of the body as the person walks.
5. Pedometer. It is similar to a passometer but it
registers directly the distance walked and not the
number of paces.
6. Odometer. It is a simple device, which can be
attached to wheel of a bicycle or any such vehicle.
It registers the number of revolutions made by the
wheel. The distance covered is equal to the product
of number of revolutions and the circumference of
the wheel.
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 Measuring tapes
Measuring tapes can be classified into 5 types
depending upon the material used in their
manufacture.
i. Liner or cloth tapes.
ii. Glass-fiber tapes
iii. Metallic tapes
iv. Steel tapes
v. Invar tap

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 Cont…
i. Linen or cloth Tapes. These tapes are made of linen or cloth. The
tape is light and handy but not very accurate. These tapes are
available in length of 10m 20m , 25m and 30m.
ii. Glass –fibre taps These tapes are similar to liner and plastic
coated tapes but these are make of glass-fibre. The tapes are quite
flexible, strong and non-conductive.
iii. Metallic tapes. These tapes are similar to liner tapes but are made
of water proof fabric or glass fibre in which metallic wires are
interwoven.
iv. Steel tapes. The steel tapes are more accurate than metallic tapes.
The steel tapes are made of steel or stainless steel strip.
v. Invar Tapes. Invar tapes are made of an alloy of steel (64%) and
nickel (36%) which has a very low coefficient of thermal
expansion.
 Invar tapes are used for linear measurements of very high
precision. prepared By Afework Legesse
 Instrument for taping
• In addition to a tape, the following small
instruments and accessories are required for
the determination of the length of a line.
Arrows (Chain pins)
Pegs
Ranging rods
Plumb bobs

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Three kinds of Errors in measurements
1. MISTKES- mistakes are errors that arise from inattention,
inexperience, carelessness and poor judgment or confusion in
the mind of the observer.
2. SYSTEMATIC ERRORS- they always follow some definite
mathematical or physical law, and a correction can be
determined and applied.
3. RANDOM (ACCIDENTAL) ERRORS:- are those which
remain after mistakes and systematic errors have been
eliminated and are caused by a combination of errors.
• Accidental errors represent the limit of precision in the
determination of a value, i.e., due to lack of perfection in the
human eye.

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 Which type of error?
A. In measuring a distance with a scale tape
marked in centimeters, one has to estimate a
distance of 5mm as 6mm or 4mm because the
eye cannot judge the exact division.
B. The error in the length of the steel tape due to
change in temperature

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 Taping corrections
1 Correction for standard
• A steel tape will normally be provided with standardizing data,
for example it may be designated as 30m long under a tension
of 50N at a temperature of 20c0 when laid on the flat. With use
the tape may stretch and it is imperative that the tape is
regularly checked against a reference tape kept specifically for
this purpose.
Ca=C*L
l
where ca = correction for absolute length
C = correction be applied the tape or the difference b/n
standard and the actual length of tape
l = nominal or designed length of the tape
L= measured length.
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 Example
• A distance is measured with a 50m steel tape and is found to
be 739.56m .Later the tape is standardized and is found to have
an actual length of 50.05m. Calculate the corrected length of
the measured distance.
Solution:
Ca=C*L
l
Given: L = 739.56m, l’ = 50.05m, l= 50m
C= 50.05 -50.00 = 0.05m
Required: absolute length
Ca = 0.05*739.56 = +0.74m
50.00
• Corrected (Absolute) Length = measured Length + Correction
for absolute length = 739.56 + 0.74m
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 2 Correction for tension (Pull correction)
If the pull applied to the tape during measurement, is more than the pull at
which it was standardized, its length increases and hence the measured distances
become less than actual. Correction for tension is therefore positive. On the
other hand, if the applied pull is less, its length decreases and consequently the
measured distance become more. The correction for tension is negative.
The ratio of stress & strain which is known as young’s modulus of the elasticity
of the material (E), E=stress : strain
stress = P/A , strain = l/L l=?
If the tape is of correct length under a standard tension and it is under a different
tension the correction which should be applied is
Cp= ( P-Ps)* L
AE

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 Cont…
where P= is the tension applied in the field in Newton (Kg)
Ps= is the standard tension. (pull)
A =is the cross sectional area of the tap
E =is Young’s modulus for the tape material and
L =is the measured length.
l = elongation of the tape
Note: the sign of the correction takes that of quantity (P-Ps)

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 Example
• A steel tape 20m long standardize with a pull of 100N was
used for measuring a base line. Find the correction for tape
length if the pull exerted during measurement was 160N. Take
cross sectional area of the tape and young’s modulus of
elasticity of the tape as 5.089x10-6m2 and 2.11 x 106 N/m2.
Solution
Cp = ( 160 – 100) x 20 = 0.118m (additive)
(5.089x10-6x 2.11 x 109)

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 3 Correction for sag
• For very accurate work the tape can be allowed hang in
catenary, free of the ground, between suitable supports.
In the case of a long tape intermediate supports can be
used to reduce the magnitude of the correction.
• If the tape has been standardized on the flat the
correction that should be applied to reduce the curved
length to the chord length is
sag correction= _ LW2
24 p2
Where W is the weight of the tape per unit length
L is the observed length and
P is the tension applied in the field.
Note: - If the tape in used on a plane surface, which can
be considered, flat then no Correction is applicable.
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 Example
Calculate the sag correction for a 300m steel tape
under a pull of 10N if the weight of the tape
was 0.17 N/m.
solution
Cg = w2 L = - (0.17) 2 (300) = -0.003m
24P2 (24) (10)2

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 4 Correction for slope
• In surveying it is essential that horizontal
lengths are determined. Thus length L
measured on the slope must be reduced to its
equivalent plane length -L(1- cosine ). The
correction to be applied is
Cs= __ h2 OR Cs= (L2-h2) 1/2-L
2L
Cs= -L(1- Cos ) or Cs= - 2L sin2 Ө/2
where h = elevation difference
L = measured slope distance
Cs = Correction for slope
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 Example
• A distance measured with a hundred meter
steel tape along an uneven ground and found to
be 238. 40m. if the elevation difference b/n the
end pts is 2.75m (or ,the slope angle is 00
39’39”) what’s the respective measured
horizontal distance.

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 Solution
• Given – S-238.40m h= 2.75m (ፀ= 00 39’39”)
• Required - H distance
Slop corr.
(1) Cs = h2 = ( 2.75)2 = 0.02
2S (2x238.40)
Or
(2) Cs = S(1 – Cos 00 39’39”)
= 238.40 (1- 00 39’39”) = 0.02M
H = S - Cs = 238.40 – 0.02 = 238.38m

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 5, Correction for temperature
• If a tape is used at a field temperature different from the
standardization temperature then the correction is
• (T-Ts) L, where is the coefficient of thermal
expansion of the tape material
 T is the field temperature
 and Ts is the standardization temperature.
•  is the coefficient of thermal expansion of the tape
material
steel: 0.0000115 m/(m C)
invar: 0.000001 m/(m C)
• Note:- The sign of the correction takes the sign of (T-
Ts).
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 Example
• A survey line was measured with a 50m tape and
found to be 135.76m at an average field temperature
of 250C. Find correction for temperature if the
standard temperature of the tape is 100c. Take x of the
tape 1.15 x 10-5 /C0
solution

Ct = (T-Ts) L = 1.15 X 10-5 ( 250C-100C) X 135.76m

= 0.023m

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 6, Correction to mean sea level
In the case of long lines the relationship between
the length measured on the ground and the
equivalent length at mean sea level has to be
considered.
In the measured length is Lm and the height of
the line above datum is H then the correction
to be applied is
correction M.S.L. = _ L* h
R

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 M.S.L.

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 7, Correction for Alignment
• If the survey line is not accurately ranged out the
error due to misalignment occurs. The measured
distance is always greater than the correct distance
and hence the error in positive and the correction is
negative.
• The correction is calculated as that for slope, where is
the distance by which the line is out the error due to
misalignment also occurs if there is some obstruction
on the survey line, and it becomes necessary to follow
a part consisting of two straight lines.
• Cm = -[l1 (1-cosӨ1)+ l2 (1-cosӨ2)]
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 Cont…
• If the end stations A and B are not inter visible,
the included angle ( ) at C can be measured
accurately with a theodolite. The correction is
then given by

• AB = (AC2+BC2-2AC.Bc cosα)1/2

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 Activity 1
The standardized length of the tape was known to be 20.015m at
250c and 50N tensions whose nominal length is 20m. The tape
is used to measure a base line suspending it b/n supports and
the following measurements were recorded:-
• Measured length = 93.421m
• Elevation difference b/n the two extreme ends = 4.482m
• Mean temperature = 230C
• Cross – Sectional area of the tape 1.7mm2
• Tension applied = 25N
• Weight of the tape = 3.4N
 Calculate (with millimeter accuracy) the corrected horizontal
distance of the base line if Young’s modulus of the tape (E)
material is 2x105N/mm2 and the coefficient of thermal
expansion of the tape (prepared
) is 1.12x10 -5/CO
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 Solution
1) Standardization Correction:-
Ca = (l'- l) L = (20.015 – 20)* 93.421M =+0.070m
l 20
2) Slope Correction:-
Cs = h2 = (4.482)2 = -0.108m
2S (2x93.421)
3) Correction for pull:-
Cp = (P - Po) L = (25N – 50N) * 93. 421 = -0.007m
AE (1.7x2x105)
4) Correction for sag
Cp = - W2L = - 3.42 x 92.421 = - 0.072m
24p2 24 x 252
5) Correction for temperature:- Ct = (Tm – To) L
= 1.12x10-5 (230c – 250c) 93. 421m = -0.002m
Total Correction = Ca + Cs + Cp + Cg + Ct = 0.070 – 0108 – 0.007 – 0.072 - 0.002
Corrected length = Measured length + Total Correction = 93.421 + (- 0.119)
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= 93.302m
 Activity 2
• A survey line was measured with a tape,
believed to be 20m long, a length of 284.62m
resulted. On checking, the tape was found to
measure 19.95m long.
a) What was the correct length of the line ?
b) If the line lay on a slope of 1 in 20 what would
be the reduced horizontal length used in the
plotting of the survey?
c) What reading is required to produce a horizontal
distance of 15.08m between two site pegs, one
being 0.66m above the other?
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 Solution
A, a length of 20m would be booked overall, but actually only a length of
19.95m has been covered.
20m= 284.62m
19.95= ?
Correct length of line = 19.95x284.62= 283.91m
20
B, A slope of 1 in 20 implies that there is a chang in height of 1m over each
20m in length horizontally
tan=1
20
=tam –1 1 = 2052’
20
Horizontal dist= The required length =283.91xcos 20 52’
283.56
• Note that horizontal distances are required when plotting survey lines
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C, the following reading is required

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 Activity 3
• A steel tape was exactly 20m long at 20⁰c when
supported through out its length under a pull of 5 kg.
A line measured with this tape under a pull of 16kg
and at a mean temperature of 32 ⁰c ,was found to be
680m long. Assuming the tape is supported at every
20m, find the true length of the line.
Given that:
i. Cross-sectional area of tape=0.03cm2
ii. E=2.1*106kg/cm2
iii. α= 11*10-6/ ⁰c
iv. Weight of tape= 600g
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 Solution
• Temperature correction =0.00264m
• Pull correction=0.00349m
• Sag correction=-0.00117m
Total correction= 0.00264+0.00349-0.00117
= 0.00496m
Actual length of tape L’=20.00496m
True length of line= L’* measured length
L
20.00496*680 =680.169m
20
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 Accuracy and Precision
• The terms accuracy and precision are constantly used
in surveying, yet their correct meanings are a little
difficult to grasp.
• Accuracy refers to the degree of perfection obtained
in measurements. It denotes how close a given
measurement is to the true value of the quantity.
• Precision is the closeness of one measurement to
another. If a quantity is measured several times and
the values obtained are very close to each other, the
precision is said to be high.

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 Cont…
• In measuring distance, precision is defined as the
ratio of the error of the measurement to the distance
measured and it is reduced to fraction having a
numerator of unity.
• Example:- If a distance of 4200 m is measured and
the error is later estimated to equal 0.7m, the
precision of the measurement is 0.7/4200 = 1/6000.
This means that for every 6000 m measured, the error
would be one m, if the work were done with this
same degree of precision.

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 AREAS
Definition: - The unit of measurement for an area is in
square meter (m2).
• This term is mostly encountered in determining the
area to be
• Excavated
• Compacted
• Surfaced
• 1m2 is the area of a square having sides whose length is
1m. Consequently every unit of length can be
converted into an area if it is multiplied by itself.

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 Methods of computing area
• There are many methods of measuring area.
But in this training we will see the following
1. Geometric method
2. Coordinate method
3. Meridian distance method(MD)
4. Double meridian distance method

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1, Geometric method

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 2, Coordinate method
• In this method independent coordinates of the
points are used in the computation of areas.
• To avoid negative sign, the origin O is chosen
at most southerly and westerly point.
• Total area of the traverse ABCD can be
calculated as follows.
[A1-A2]
2

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Y

C
A

D
X

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• Two sums of products should be taken
1. product of all adjacent terms taken down to the
right
A1 = XAYB + XBYC + XCYD + XDYA
2. Product of all adjacent terms taken up to the right
A2 = YAXB + YBXC + YCXD + YDXA
• The traverse area is equal to half the absolute
value of the difference between these two sums.
• In applying the procedures, it is to be observed
that the first coordinate listed must be repeated at
the end of the list.
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 Activity 1
• Calculate the area of closed traverse of
polygon ABCDEF.
Point X Y
A 500.000 1000.000
B 416.693 578.866
C 1047.169 395.856
D 1297.375 564.653
E 1330.387 650.165
F 861.433 1090.090
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 Solution
• XAYB = 500.00 * 578.866 = 289433
• XBYC = 416.693 * 395.856 = 164950.4242
• XCYD = 1047.169 * 564.653 = 591287.1174
• XDYE = 1297.375 * 650.165 = 843507.8169
• XEYF = 1330.387 * 1090.090 = 1450241.565
• XFYA = 861.433 * 1000.000 = 861433.0
ΣA1 = 4200852.923m2

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• YAXB = 1000.000 * 416.693 = 416693
• YBXC = 578.866 * 1047.169 = 606170.5304
• YCXD = 395.856 * 1297.375 = 513573.678
• YDXE = 564.653 * 1330.387 = 751207.0107
• YEXF = 650.165 * 861.433 = 560073.5864
• YFXA = 1090.090 *500.00 = 545045.0
ΣA2 = 3392762.806m2
• Area = [ΣA1- ΣA2 ]
2
= 808090.1175
2
Area = 404045.059sq.m
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 Activity 2
• Find the area enclosed by the traverse
ABCDEFA whose coordinates are the
following
A (0,0) B( 300,100) C(450,350) D(425,600)
E(200,600) F(-100,350)

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 Measurement of Angle
• Measurement of angles is basic to any survey
operation. When an angle is measured in a
horizontal plan it is horizontal angle when
measured in a vertical plane it is vertical angle.
In surveying, the direction of a line is
described by the horizontal angle that it makes
with a reference line or direction. A theodolite
is an important instrument used for measuring
horizontal and vertical angles in surveying.

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 Different Types of Horizontal Angles
measuring methods
Horizontal angles can be classified as
• Interior angles,
•Exterior angle or
• Deflection angles.
Interior angles can be clockwise when the direction of turning is
clockwise, anticlockwise when the direction of turning is
anticlockwise. Similarly deflection

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 …..cont
B B
E
A C A Clockwise
C

E D E D
A B
Anticlockwise left deflection angle
Closed Polygon-
instrument station
Clockwise C

Right deflection C
B
angle
Deflection
angle.
A
Different type of angles: (a) Closed Polygon-instrument station
A,B,C,D and E all angles measured clockwise
b) Closed polygon –instrument station A,B,C,D and E all angles measured
anticlockwise
c) Deflection angle. prepared By Afework Legesse
 Direction of a line
Direction of a line is the horizontal angle from a reference line called the Meridian.
There are Four basic types of Meridians.

1) Astronomic meridian: it is an imaginary line on the earth’s surface

passing through the north-south geographical poles.
2) Magnetic meridian: it is the direction of the vertical plane shown by a
freely suspended magnetic needle.
3) Grid meridian: a line through a point parallel to the center meridian or Y-
axis of a rectangular coordinate system.
4) Arbitrary meridian: an arbitrary chosen line with a directional value
assigned by the observer. These are explained graphically.

Magnetic variation or
Astronomic North declination

Grid North Mapping

angle

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 Bearing
Bearing of a line is measured from the north or south terminus of a
reference meridian. It is always less than 90o and is designated by the
quadrant in which it lies as shown in the figure.N From Athe figure it can be
seen that D
Bearing of OA=N40o E 40o
40o
OB=S25oE IV I
OC=S30oW W E

OD=N45oW III II
o o30 25

B Quadrennial
S Bearing

Since bearing is with reference to N.S line angles are measured clockwise
in the 1st (NE) and 3rd (S.W) quadrant. It is measured anticlockwise in 2nd
and 4th quadrants (NW and SE).

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 Azimuths
oAzimuths are angles measured clockwise from any reference meridian. They
are measured from the North and vary from 0o to 360o and don’t require
letters to identify their quadrant. The figure shows the azimuths of different
lines whose bearings are given
N
A
D

40o
45o
W E
0

25o
Azimuth or whole circle bearing
30o
C B
Azimuth of OA = 40o
S
OB = 180o – 25 = 155o
OC = 180o + 30o = 210o
OD = 360o – 45o = 315o

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 Measurement of vertical angles
• Vertical angles are measured on a true vertical plane for computing the height.
slope distance, elevations..etc of the points or object on the surface of the earth.
• The following sketches are self explanatory to measure angles at different axis of
reference.

+90o 0o 180o

0o 0o 90o 270o 90o 270o

-90o 180o 0o

Angles from the Zenith angles Nadir angles

Horizon

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 Types of vertical angle
1. Angle of elevation &
2. Angle of depression
• The angles measured above the horizon
are called angle of elevation.
• The angles measured below the horizon
are called angle of depression

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 Errors in angle Measurement
The errors in angle measurement occur when the measured angles are not in
true horizontal plane and vertical plane , It is possible due to various
Instrumental, natural and personal reasons ,which can be tabulated as below

Sl.
Type of error Cause of error Correction
no.

1) Vertical axis of the instrument is

not parallel to the plumb line Instrument need to
2) Line of collimation is not under go proper
1. Instrumental perpendicular to horizontal axis of
the instrument temporary and
3) Horizontal axis is not permanent adjustments
perpendicular to true vertical axis

Imperfect bisection to the object and Under proper field

2. Personal
temporary adjustments. training
Sun/wind/vibrations/electromagnetic Use umbrella / avoid
3. Natural
interference ..etc bad weather conditions
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 Measuring distance with stadia methods
• Distance can be measured indirectly by the use of optical
instruments in conjunction with leveling staff (rod). The
measurements are performed quite rapidly and are
sufficiently accurate for many types of surveying
operations like Stadia or Tachometry.
• The word tachometry means fast measurement".
• The telescope of the Theodolite usually contains three
horizontal lines, which are attached on the eyepiece of a
telescope, known as cross hairs. The upper (U) and
lower (L) cross hairs are called Stadia hairs. The staff
readings are taken form staff held vertically
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 Cont…
• The distance on horizontal sighted from the center of the
instrument are given by
• HD=KS
• Where H = horizontal distance on a level
surface
• K = telescope constant (100)
• S=: staff intercepts (U - L)

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Staff readings

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Leveling

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Outline:
 Definition
 Purpose
 The Equipment
 Differential Leveling
 Definition of Terms
 Computation of Elevations
 Common Mistakes
 Suggestions for Good Leveling
 Leveling Errors
 Profile Leveling
 Worksheet

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Definition: Leveling
 The determination of elevations is called
leveling

 Leveling is the process by which differences in

height between two or more points can be
determined.

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Definition: Datum
 Datum - This is an arbitrary level surface to
which the heights of all points are referred.

 This may be the National Datum or local datum

point established on a construction site.

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Definition: MSL
 Mean sea level (MSL) is the average (mean)
height of the sea between High and Low tides

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Definition: RL
 Reduced Level (RL) – A distance recorded as
a Height Above or Below the DATUM. This heig
ht is in metres

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Definition: BM
 In surveying a benchmark is specifically any
permanent marker placed by a surveyor with a
precisely known vertical elevation (but not necess
arily a precisely known horizontal location).

 Designed to be used for many projects.

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Definition: TBM
Temporary benchmark (TBM):

 A Benchmark usually placed for a particular

project.

 Not designed to be a reference for other

projects or for long term use

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Definition of Terms
 Backsight (BS) - The first sight taken after setting the
instrument up
 Foresight (FS) - The last sight taken before the instru
ment is moved)
 Height of instrument (HI) - the elevation of the line
of sight of the telescope
 Intermediate sight (IS) - Any sighting that is not a
back sight or fore sight
 Line of Collimation – Imaginary line that passes throu
gh leveling instrument at Cross-Hairs
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Definition of Terms
 Height of collimation: Elevation of line of sight obtained by
adding BS reading to the elevation of point on which BS reading is
taken

 Focusing: Aiming the telescope on rod and then focusi

ng it for a clear vision of rod.
 Turning Point (TP): Also called Transfer Point (TP) or
change point. Selected and used to transfer elevation b/n
BM. On every turning point one FS and one BS are taken.

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Purpose
 To find the relative heights of things,
 To find the absolute height of an object,
 To provide heights or contours on a plan,
 To provide data for road cross-sections ,
 To provide volumes of earthworks,
 To provide a level or inclined surface in the setting out
of construction works,
 etc…

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 Leveling Instruments

• Levels are categorized in to four groups.

1) Dumpy levels
2) Tilting levels
3) Automatic levels
4) Digital levels

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• Dumpy levels
In dumpy level, the line of sight is perpendicular the vertical axis. Once the
instrument is leveled the line of sight becomes horizontal and the vertical
axis becomes truly vertical provided the instrument in adjacent.
• Tilting levels
It has the telescope that can be tilted about a horizontal axis. This design
enables the operator to quickly and accurately center the bubble and brings
the line of sight in to a horizontal plane.
• Automatic levels
One of the most significant improvements in leveling instrumentation has
been automatic level or self-leveling levels. It has an internal
compensatory that automatically makes horizontal the line of sight and
maintains the position through the application of the force of gravity. As
soon as the instrument is leveled by a means of a circular bubble, the
movable component of the compensatory swings free to a position that
makes the line of sight horizontal.

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The Equipment: The Level
 Has a set of cross-hairs
 Can be turned through 360° horizontally
 Consist of a high-powered telescope
 Attached to a spirit or bubble level that keeps the line
of sight of the telescope horizontal

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The Equipment: The Tripod
 The Tripod:- A fully adjustable 3-legged stand
on which the level sits.

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The Equipment: The Staff
 A measuring stick, usually 4m tall, and clearly
marked in divisions of 10mm.

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Reading the E Staff

• Each “E” is 50mm

• Each Part of the E is 10mm

• Millimeters are interpolated

• Staff is read to the millimeter

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The Equipment: Bubble
 Keep the staff upright
 Any tilt will disturb your readings

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The Equipment: Automatic level

1. Base Plate
2. Horizontal Circle
3. Eyepiece
4. Circular Bubble
5. Sighting Pointer
6. Objective Lens
7. Focusing Knob
8. Fine Motion Drive
9. Footscrew
10.Bubble Mirror prepared By Afework Legesse
 The main parts of level
1 foot screw
2 Horizontal Slow Motion
3 focus Control

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Methods of leveling can be broadly classified as:

i. Simple leveling
ii. Differential leveling,
iii.Profile leveling
iv.Cross sectional leveling
v. Reciprocal leveling and etc

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 Direct or Simple Leveling
• Let us now examine in turn each of the above
types of levelling. Let us assume that the
elevation of station A from a given datum is
known and that the elevation of a second point
B is to be determined from the same reference
datum. If two points are so situated that they
are visible from a single set up of the level, the
instrument is set up approximately mid-way
between the two points.
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prepared By Afework Legesse
 simple Leveling

Δh = BS - FS

BS = 6.32 ft HI = 106.32 ft FS = 3.10 ft

Point B

Point A
Elevation = 103.22 ft

Starting point
(elevation 100.00 ft) By Afework Legesse
prepared
Simple Differential Levelling
Differential Leveling

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Differential Leveling
ΣΔh = AElev – Belev

= ΣBS – ΣFS
BS

FS
FS
BS FS BS
BS FS

A
TP # 1 TP # 2 B
TP # 3

Compound Differential Levelling

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Station Chainage BS IS FS hi RL Remark

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Computation of Elevations

BS
12.64

BM1
1. BS + Elevation = HI
Elevation 100.00

Point BS HI FS Elevation
BM1 12.64 112.64 100.00

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Computation of Elevations

BS FS
12.64 3.11

BM1
TP1 2. HI - FS = Elevation
Elevation 100.00

Point BS HI FS Elevation
BM1 12.64 112.64 100.00
TP1 3.11 109.53

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Computation of Elevations

BS
BS FS 10.88
12.64 3.11

BM1
TP1
Elevation 100.00

Point BS HI FS Elevation
BM1 12.64 112.64 100.00
TP1 10.88 120.41 3.11 109.53

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Computation of Elevations

BS FS
BS FS 10.88 2.56
12.64 3.11

BM1 TP2

TP1
Elevation 100.00

Point BS HI FS Elevation
BM1 12.64 112.64 100.00
TP1 10.88 120.41 3.11 109.53
TP2 2.56 117.85

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Computation of Elevations

BS
9.72
BS FS
BS FS 10.88 2.56
12.64 3.11

BM1 TP2

TP1
Elevation 100.00

Point BS HI FS Elevation
BM1 12.64 112.64 100.00
TP1 10.88 120.41 3.11 109.53
TP2 9.72 127.57 2.56 117.85

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Computation of Elevations

BS FS
9.72 3.10
BS FS
BS FS 10.88 2.56
12.64 3.11 BM2

BM1 TP2

TP1
Elevation 100.00

Point BS HI FS Elevation
BM1 12.64 112.64 100.00
TP1 10.88 120.41 3.11 109.53
TP2 9.72 127.57 2.56 117.85
BM2 3.10 124.47
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Computation of Elevations
ΣΔh = AElev – Belev

Change in elevation: = ΣBS – ΣFS

summation of the BS and the FS then subtract

Point BS HI FS Elevation
BM1 12.64 112.64 100.00
TP1 10.88 120.41 3.11 109.53
TP2 9.72 127.57 2.56 117.85
BM2 3.10 124.47
+33.24 -8.77

Change in elevation = 33.24 -8.77 = 24.47

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Computation of Elevations
Example:- What are the elevations of points TP1 and TP2?

BS FS
1.27 4.91
BS FS
2.33 6.17
BM
Elevation 356.68

TP1
TP2

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Computation of Elevations
Answer

BS FS
1.27 4.91
BS FS
2.33 6.17
BM
Elevation 356.68

TP1
TP2
Point BS HI FS Elevation
BM1 1.27 357.95 356.68
TP1 2.33 355.37 4.91 353.04
TP2 6.17 349.20
+3.60 -11.08 -7.48
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 Profile Leveling

The process of determining the elevation of points at short measured intervals

along a fixed line is called profile levelling. The need of profile levelling arises
during the location and construction of highways, railroads, canal, and sewers.
The following figure shows the plan, and sectional elevation of a road way along
which a line of level is being taken. The figure also explains the different terms
used in connection with differential leveling.

B.M
P Q
X X X X X X X
R.L.
A B C D E F G
100.545

Back sight 1.815

= .515m. 1.525 1.095

1.605 1.655
B.M. 1.515
R.L.=
100.545 P Q
1.645
Fall between 1.715
B.M. and A.
A B C D E F G 132
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Profile Leveling
 To collect data about topography along a refere
nce line

 Mainly to compute volumes of cut and fill for a

proposed linear structure, such as:

o highways,

o railroads,

o transmission lines,

o canals.

 Profile leveling establishes a side view or cross

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Profile Leveling

Profile leveling establishes a side view or cross section

al view of the earth’s surface
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Profile Leveling

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 Cross sectional leveling
Cross-section is a vertical section taken normal to the
direction of the proposed center line of an engineering
project. X-sections are run at right angles to the
longitudinal profile & on either side of it for the purpose
of lateral out line of the ground surface.
They provide the data for estimating quantities of Earth
work & for other purposes. The x-section are numbered
consequently from the commencement of the center line
& are set out at right angles to the main line of section,
the distance are measured left & right from the center
line of the road. The length of the cross-section depends
up on the nature of the work and relief type.
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Chain Distance BS IS FS hi RL Rema
age L C R rk
0+000 0
2
4
6
2
4
6
0+030 0

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Typical Cross-section

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 Reciprocal leveling
• When a line of levels crosses a broad body of water it is
impossible to balance the back sight and foresight distances, it
is necessary to take sights much longer than permissible.
Under such a measurement errors due to curvature and
refraction become significant. To obtain the best results we
should have to use the procedure termed as reciprocal leveling.
• Elevation of A = (a - b) + (c - d) + Elevation of BM1
2
Example
If a = 1,442m, b= 1.911m, c= 1.768m, d= 2.325m
and Elevation of BM1 = 1980.40m
Elevation A = 1980.4 – 0.517 = 1979.887m

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•The elevation of survey point A is to be determined by leveling from BM1. At a
set up near BM1, a back sight is taken on BM1a foresight on A. the difference in
elevation is computed as (BS–FS). Next the level is set up near point A. assuming
that atmospheric refraction remains constant during the time between the two set
ups, the correction differences in elevation is computed as the mean of the two
measured differences.

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 Level Book
Instead of writing the readings in a sketch and giving suitable descriptions, the whole
process of leveling is systematically shown in a level book and reduced levels of
different points found out. There are two methods of reducing levels.
(i) Rise and fall method, and
(ii) Height of collimation method.
Complete bookings and reductions in the two methods are given in the following
table
Back- Inter- Force- Rise Fall Reduced Distanc Remark
Sight sight sight level e in m
0.515 100.545 Benchmark
1.525 1.010 99.535 0 Staff Stn. A
1.095 0.430 99.965 30 Staff Stn. B
1.645 0.550 99.415 60 Staff Stn. C
1.815 1.515 0.130 99.545 90 Staff Stn. D
(Change point)
1.715 0.100 99.645 120 Staff Stn. E
1.605 0.110 99.755 150 Staff Stn. F
1.655 0.050 99.705 180 Staff Stn. G
∑=2.33 ∑ 3.170 ∑ ∑
0 0.770 1.610
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Check: ∑ Back-sight - ∑ Fore-sight
= (2.330) – (3.170) = -0.840
∑ Rise - ∑ Fall
= (0.770) – (1.610) = -0.840
Last R.L. – 1st R.L.
= (99.705) – (100.545) = -0.840
Rise and Fall Method
Each reading is entered on a different line in the applicable column, except at
change points where a fore-sight and a back-sight occupy the same line. This is
to connect the line of sight of one setup of the instrument with the line of sight of
the second setup of the instrument. From the above figure it can be seen that they
are not at the same level. R.L. of change point D is obtained from the first line of
sight by comparing intermediate sight 1.645 with foresight 1.515, i.e. a rise of
0.130m. For the R.L. of next point E, back sight 1.815 is compared with
intermediate sight 1.715, i.e. a rise of 0.100m. At the end of the table arithmetic
checks are shown.
The checks are:
∑ Backsights - ∑ Foresights = ∑ (Rises) - ∑ (Falls)
= Last R.L. – First R.L.
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 Back- Inter- Force- Ht. of Reduced Distance Remark
Sight sight sight collimation level
0.515 101.060 100.545 B.M.
1.525 99.535
1.095 99.965
1.645 99.415
1.815 1.515 101.360 99.545
1.715 99.645
1.605 99.755
1.655 99.705
∑=2.330 ∑ 7.585 ∑ 3.170

Check: ∑ Backsights - ∑ Foresights = 2.330 – 3.170

= - 0.840
Last R.L. – First R.L. = 99.705 – 100.545
= - 0.840
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 Trigonometrical levelling
 Trigonometrical levelling is done mainly to find the height of the
buildings/chimneys/towers or any such remote , inaccessible objects. This principle is
also being used in general topographical survey/Engineering projects.

 In this method distances and vertical angles are measured to compute the reduced levels,
heights and other parameter as required by trigonometrical relation.

 Staff reading A
Chimney

Bench mark
RL=100
D
RL of Chimney top = RL of BM + Staff reading A + H H = DTanӨ

RL of Chimney top = RL of BM + Staff reading


A + DTanӨ

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Common Mistakes
 Misreading the rod - reading 3.54 instead of 3.45

 Moving the turning point - use a well–defined TP

 Leveling rod not fully extended or incorrect length

 Level rod not vertical

 Level instrument not level

 Instrument out of adjustment

 Environment - wind and heat

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Suggestions for Good Leveling
 Anchor tripod legs firmly

 Check the bubble level before and after each reading

 Provide the rod person with a level for the rod

 Always start and finish a leveling run on a BM and close the loops

 Keep BS and FS distances as equal as possible

 Keep lines of sight short (normally < 50m)

 Never read below 0.5m on a staff (refraction)

 Use stable, well defined change points

 Staff should be set up vertically

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Leveling Errors

 There are a large number of potential sources of errors

in leveling. Many of these are only significant for precis
e leveling over long distances. For the short segments
of leveling to nearby benchmarks there are only four w
orth mentioning:

o Collimation Error

o Error due to Earth Curvature

o Error due to Parallax Error

o Error due to Refraction

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Worksheet-3
Some readings are omitted from the following level book. The elevati
on of BM-A is 2.255m less than the elevation of BM-B. Compute the
missed data and calculate the elevation of all stations. Assume that
there is no apparent error in measurements.

Station BS IS FS Rise Fall Elevation

BM-A 0.685 ?

X 1.105 ? 1.025 ?

Y ? 2.435 ? ?

Z 1.680 0.650 2.395 ?

S 1.145 ? 610.760

T 1.500 ? 0.561 ?

U ? 1.325 ? ?

K 1.640 2.225 ? ?

BM-B 0.895
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Q&A

?
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Control
surveying
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 General objectives
At the end of this you will be able to
i. Define “traversing “ by your own words
ii. Identifying the types of traverse.
iii. Define and identify the difference between departure and
latitude.
iv. Ways of calculating departure and latitude in different cases
(when distance and azimuth are given, or when two coordinates
are given.
v. Computing the azimuth of consecutive lines by using one
reference.
vi. Compute of relative coordinates
vii. Balance a traverse.
viii. Computing of a coordinates by using traverse tables.
ix. Run different types of traverse outside in the field.
x. To solve problems regarding traverse.
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 What is control surveying ?
• A control survey provides a framework of
survey points, whose relative positions, in two
or three dimensions.
• Control Surveying – establish a network of
horizontal and vertical monuments that serve
as a reference framework for other survey
projects.

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The methods used for control surveys
are:
i. Traversing
ii. Triangulation
iii. Intersection
iv. resection
v. Trilateration
vi. Satellite position fixing (GPS)

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 i. Traversing

• Lines
• Distance
• Angles &
• Directions/ Coordinates

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 Types of Traverse
• General a traverse can be divided in to two
i. Closed traverse
A. Closed loop (ring) traverse
B. Closed route (link) connection)
traverse
ii. Open traverse

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Closed traverse

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Open traverse

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 Cont…

• Discuss the difference between

closed and open traverse.

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Measurements of Traverse angles in the field
• Angle – difference in direction of 2 lines
 Another way of explaining is the
amount of rotation about a central
point
 3 kinds of Horizontal angles: Exterior
( to right); Interior; Deflection
 To turn an angle you need
• A reference line
• Direction of turning
• Angular distance
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 Cont…
Angular Units
• Degrees, minutes, seconds (sexagesimal
system)
 Circle divided into 360 degrees
 Each degree divided by 60 minutes
 Each minute divided into 60 seconds
• Radians
 1 radian = 1/2 of a circle = 0.1592*360
= 5717’44. 8”
• Grads (Centesimal System) – now called Gon
 1/400 of a circle or 054’00” (100 gon =
90)

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 Cont…
• If traverse closes: sum of the interior angles should equal the
sum of
– (N-2)X180, (2N-4)x90
– (4N-8)x45, (8N-16)x2230’
– where N = Number of sides
• 3 angles = (3-2) 180 = 180
• 4 angles = (4-2) 180 = 360
• 8 angles = (8-2) 180 = 1080
• 25 angles = (25-2) 180 = 4140
– If an exterior angle exists, subtract it from 360 to obtain
the interior 
– Angular closure should be checked before leaving the field
– Allowable error = 1’ (n) 1/2
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 Computation of Azimuths consecutive
lines
• If azimuth of one line and included angle at
stations are given we can calculate azimuth of the
other lines
• By using the given azimuth of the first line find
the azimuth of all lines.
• If it is right hand traverse
• Az of (i+1) = Az of i + 180  -ß
• If it is left hand traverse
• Az of (i+1) = Az of i + ß - 180 
• Where ß is adjusted angle
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 Bearings Vs Azimuths
4 Point Comparison
Bearing Azimuth

1. Numeric Value 0-90 0-360

2. Method of 2 letters & number Number only
Expressing
3. Direction Clockwise & Clockwise
counterclockwise
4. Position of 0 North and South North
point
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Whole Circle Bearing

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 Cont…
• It is always very important to have your field
sketch properly oriented

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 Balancing the traverse
• Balancing the traverse means making
adjustment to remove any apparent error. For
balancing the traverse the underlying objective
is to adjust the traverse in such away that the
sum the latitude and departure should each
equal to zero in closed loop traverse. The
closing error, however it is distributed
throughout the traverse such that the above
mentioned objective is achieved this operation is
called Balancing the traverse.

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 Departure and latitude
• Departure:- The difference in x- coordinate
between two points
• Latitude:- The difference in y- coordinate
between two points
i. By using coordinates
ii. By using distance and azimuth

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 1. By using coordinates

Departure of AB = XAB= XB- XA

Latitude of AB = YAB= YB-YA
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 Activity 1

1. A Point K has a Coordinate of (400,543) and

anther S has a coordinate of (521,269)
a. Calculate departure and latitude of line
joining K to S.
b. Calculate departure and latitude of line
joining S to K.

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2, By using distance and azimuth

Departure = distance * sin of Azimuth.

Latitude = distance* cost of Azimuth.
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Angles and Determination of Direction
Rectangular Coordinates
• Totally based on computation of right triangle
North – South Movement =
Latitude = D X cos A
East – West Movement =
Departure = D X sin A
• Latitude running North are +, South are –
• Departure running East are +, West are –

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 Activity 2

• A Surveyor determines a length and

azimuth of a line ST and found that
Length = 782.3m
Azimuth = 112 11’20’’
i. determine dep. and Lat of line ST
ii. determine dep and lat of line TS

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Basic Procedure of Traversing
1. Determine Latitude and Departure
2. Sum Lat. and Departure to calc. closure
3. Obtain balanced Lat. and Dept.
(Compass Rule)
4. Determine coordinates
5. Once rectangular coordinates are known
on point, their exact location is known
with respect to all other points in the
network
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 Steps-by -steps by procedure:
 Determine directions of traverse side

 Fill in the traverse computation table

 Compute the angular error & adjust the angles

 Compute azimuth

 Compute departure and latitude

 Compute the error of closure

 Compute correction for departure and latitude

 Adjust departure and latitude

 Compute coordinates of X and Y

 Determine area enclosed

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 Determine directions of traverse side:
 A  B? (left sided traverse)
 A  E? (right sided traverse)
AZ of AB=141°45’
Station Distance (m) A= (100, 300)
A
315.62
B
E
502.43 D
C 92°20’
145°37’
176.95 A
86°26’
D
108°24’
502.06
107°8’ C
E
187.05
B
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 Fill in the traverse computation table:
Measure Adjuste Azimut Horizo Unadjuste Correctio Adjusted Coordinates
Stati d d hs nta d n
on H. angle H. angle distanc dep. lat. dep lat. dep. lat. x y
e .
A 86°26’
E
107°8’ D
B
92°20’
145°37’
C 108°24’ A
86°26’
108°24’
D 92°20’
107°8’ C

E 145°37’
B
A

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 Compute the angular error & adjust the angles:
Error = -5’
Total = (n-2)(180) e = x - xt
error = measured value – true value Corr. = -(-5’) /5
n = 5, Total = 540° = +1’Coordinates
Measured Adjusted Azimuths Horizonta Unadjusted Correction Adjusted
Station
H. angle H. angle distance dep. lat. dep. lat. dep. lat. x y

A 86°26’ 86°27’

E
B 107°8’ 107°9’ D

92°21’
C 108°24’ 108°25’ 145°38’
A
86°27’
108°24’
D 92°20’ 92°21’
107°8’ C
E 145°37’ 145°38’

B
A

539°55’ 540°
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 Compute azimuth:
AzBC = AzAB + B - 180° (left sided traverse) NB: Use Adjusted Angles
E.g. AzBC = 141° 45’ + 107°9’ - 180° = 68°54’
Measure Adjuste Azimut Horizo Unadjuste Correctio Adjusted Coordinates
Stati d d hs nta d n
on H. angle H. angle distanc dep. lat. dep lat. dep. lat. x y
e .
A 86°26’ 86°27’
141°45’
B 107°8’ 107°9’
68°54’ E D
C 108°24’ 108°25
’ 92°21’
357°19’ 145°38’
92°20’ 92°21’ A 141° 45’
D
269°40’
86°27’ 108°25’
E 145°37’ 145°38 107°9’
C
’
235°18’ prepared By Afework Legesse B
 Compute dep. & lat:
Latitude = Length*Cos(Az) LatAB = 315.65 Cos(141°45’) = -247.86
Departure = Length*Sin(Az) DepAB = 315.65 Sin(141°45’) = 195.40
Measured Adjusted Azimuths Horizonta Unadjusted Correction Adjusted Coordinates
Station
H. angle H. angle distance dep. lat. dep. lat. dep. lat. x y

A 86°26’ 86°27’
-
141°45’ 315.62 195.40 247.86
E
D
B 107°8’ 107°9’
68°54’ 502.43 468.74 180.87

C 108°24’ 108°25’ A
-8.28
357°19’ 176.95 176.76

D 92°20’ 92°21’
269°40’ 502.06
- -2.92 C
502.05

E 145°37’ 145°38’
- -
235°18’ 187.05 153.78 106.48
B
A
1684.11 0.03 0.37
539°55’ 540° prepared By Afework Legesse
 Compute the error of closure:
 You should end up where you started
o Sum of Lat’s = 0
o Sum of Dep’s = 0
 Linear Misclosure (error)
o A line connects starting and ending point
o Linear error = length of line

eDep = 0.03 eLin  2

eLat  eDep
2

eLat = 0.37  0.362  0.032  0.37

 Relative Error
o Relates error to total distance surveyed
o Expressed as 1/xxxx eLin 0.37 1
RE   
Length 1684.11 4550
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 Compute correction for dep & lat:

 Compass Rule – more common

o Assumes angles are as accurate as distances
o Proportion Lat, Dep error to length of side and total distance

Correction is the term more popularly being used to define the

magnitude of error but opposite in sign

  eLat    eDep 
Lat  Length   Dep  Length  
 Lengths   Lengths 

  0.36 
Lat AB  315.62     0.07
 1684.11 

  0.03 
DepAB  315.62     0.006  0.01
 1684.11 
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 . . . correction for dep. & lat:

Measured Adjusted Azimuths Horizontal Unadjusted Correction Adjusted Coordinates

Station
H. angle H. angle distance dep. lat. dep. lat. dep. lat. x y

A 86°26’ 86°27’
-
141°45’ 315.62 -
195.4 247.86 -.07
.01
0
B 107°8’ 107°9’

68°54’ 502.43 -
468.7 180.87 -.11
.01
4
C 108°24’ 108°25’
-8.28
357°19’ 176.95 0 -.04
176.76

D 92°20’ 92°21’
- -2.92 -.01 -.11
269°40’ 502.06
502.0
5
E 145°37’ 145°38’
- - 0 -.04
235°18’ 187.05
153.7 106.48
8
A
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1684.11 0.03 Legesse
0.37
539°55’ 540° -
-.37
 Adjust departure and latitude:
Measured Adjusted Azimuths Horizontal Unadjusted Correction Adjusted Coordinates
Station distance
H. angle H. angle dep. lat. dep. lat. dep. lat. x y

A 86°26’ 86°27’
-
141°45’ 315.62 - -
195.4 247.86 -.07
.01 195.39 247.93
0
B 107°8’ 107°9’

68°54’ 502.43 -
468.7 180.87 -.11 180.76
.01 468.73
4
C 108°24’ 108°25’
-8.28
357°19’ 176.95 0 -.04 -8.28 176.72
176.76

D 92°20’ 92°21’
- -2.92 -.01 -.11
269°40’ 502.06 -
502.0 -3.03
502.06
5
E 145°37’ 145°38’
- - 0 -.04
235°18’ 187.05 - -
153.7 106.48
153.78 106.52
8
A

539°55’ 1684.11 0.03 0.37 0 0

540° -
-.37
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 Compute coordinates of X and Y:
XB = XA + Dep AB YB = YB + Lat AB
Measured Adjusted Azimuths Horizontal Unadjusted Correction Adjusted Coordinates
Station distance
H. angle H. angle dep. lat. dep. lat. dep. lat. x y

A 86°26’ 86°27’ 100 300

-
141°45’ 315.62 - -
195.4 247.86 -.07
.01 195.39 247.93
0
B 107°8’ 107°9’ 295.39 52.07

68°54’ 502.43 -
468.7 180.87 -.11 180.76
.01 468.73
4
C 108°24’ 108°25’ 764.12 232.83
-8.28
357°19’ 176.95 0 -.04 -8.28 176.72
176.76

D 92°20’ 92°21’ 755.84 409.55

- -2.92 -.01 -.11
269°40’ 502.06 -
502.0 -3.03
502.06
5
E 145°37’ 145°38’ 253.78 406.52
- - 0 -.04
235°18’ 187.05 - -
153.7 106.48
153.78 106.52
8
A 100 300

539°55’ 1684.11 0.03 0.37 0 0

540° -
-.37
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 Determine area enclosed:
Measured Adjusted Azimuths Horizontal Unadjusted Correction Adjusted Coordinates
Station distance
H. angle H. angle dep. lat. dep. lat. dep. lat. x y

A 86°26’ 86°27’ 100 300

-
141°45’ 315.62 195.4 247.86 - -
-.07
0 .01 195.39 247.93

B 107°8’ 107°9’ 295.3

52.07
9
E502.43
68°54’ 468.7 180.87
-
.01
-.11 D
468.73
180.76
4

C 108°24’ 108°25’ 764.1 232.8

92°20’ 2 3
145°37’
-8.28
A 357°19’ 176.95 176.76 0 -.04 -8.28 176.72

D 92°20’ 92°21’
86°26’ 755.8 409.5
108°24’ 4 5
- -2.92 -.01
107°8’ C
269°40’ 502.06 -.11 -
502.0 -3.03
502.06
5

E 145°37’ 145°38’ 253.7 406.5

8 2
- -
235°18’
B187.05 153.7 106.48
0 -.04
-
153.78
-
106.52
8

A 100 300
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539°55’
1684.11 0.37 - - 0 0
540°
 Coordinates
• Computation of relative coordinates
• XB = XA + dep AB YB = YA + lat AB
• XC = XB+ dep BC YC = YB + lat BC

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 Area calculation by Coordinate method

• In this method independent coordinates of the points

are used in the computation of areas.
• Total area of the traverse ABCD can be calculated as
using the computed coordinates

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 Activity 3
• Suppose the coordinates of point A are
(1000.2, 2341.32) and departure and latitude of
line AB are 300.32 and 543.2 respectively,
determine the coordinates of point B.

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 Activity 4
1. The distance between two points is 345.98m and the
bearing of a line joining these two points is S4023’12”E
. If the coordinate of the first point is (123.45, 567.89)
determine the coordinate of the second point.
2. The coordinate of point A is (567.98, 352.87) if departure
and latitude of line AB are 123.76 and -132.51
respectively, determine the coordinate of point B.
3. The distance between two points is 345.98m and the
bearing of a line joining these two points is
S4607’12”W. If the coordinate of the first point is
(123.45, 567.89) determine the coordinate of the second
point.
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 Project 1
• Form a group of five (5) and do the following outside in
the field..
i. Select six points which are inter visible to each other and
mark with paint (A,B,C,D,E,F)
ii. Measure the horizontal distance between points. And
record it
iii. Using a Theodolite measure the interior angles of a
polygon. And record it.
iv. Sum up and compare with the nominal (theoretical) value
v. Using surveyor compass, determine the azimuth of line
AB.
vi. Taking the coordinates of A (1234.567,7654.321)
vii. Calculate the coordinates of stations B,C,D,E & F.
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Thank you!
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Any Question?

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