SUTP101
SURVEYING: THEORY AND PRACTICAL
MANGOSUTHU UNIVERSITY OF
TECHNOLOGY
Learner Guide
FACULTY OF ENGINEERING
DEPARTMENT: Civil Engineering and Surveying
Qualification: Diploma in Civil
SUBJECT: SURVEYING THEORY AND
PRACTICAL
Subject Code: SUTP101
Revised by PN Mpantsha
November 2023
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Chapter 1
INTRODUCTION TO SURVEYING
1.1
Surveying
Surveying is the art and science of making field measurements on or near the earth
surface of the earth.
Surveying field measurements include:
•
•
•
Horizontal and slope distances
Vertical distances
Horizontal and vertical angles
In addition to taking measurements in the field the surveyor can derive related
distances and directions through geometric and trigonometric analysis.
1.2
Geomatics
Geomatics, is a relatively newly coined word, derived from the word geo-, meaning
of the earth, and -matics, a derivative of informatics, meaning the science of
disseminating information.
Geomatics therefore is reformed surveying encompassing modern measurements
science, land information sciences and spatial data management. It is a discipline
with focus on information dissemination as contrasted to data collection for which
surveying was known for centuries.
1.3
Types of Surveying
Surveying can be distinguished between plane surveying and geodetic surveying.
1.3.1. Plane Surveying
Type of surveying that ignores the curved surface of the earth and assumes
the measured earth surface as a plane. The vast majority of engineering and
construction projects are limited in geographic size and therefore fall in this
category of surveying.
1.3.2. Geodetic Surveying
Surveys that cover large geographic areas for which corrections need to be
made to the field measurements in order to reflect the curved (ellipsoidal)
shape of the earth’s surface. Examples include provincial boundary surveys.
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SURVEYING: THEORY AND PRACTICAL
Branches of Surveying
Various branches of surveying/ geomatics exist depending on where they are
applied. These include:
1.4.1 Engineering surveys
Surveys preparatory to, or in conjunction with the erection of engineering
works such as roads, railways, dams, tunnels, storm water and sewerage
works, and construction works generally including the calculation and placing
of pegs in the field
1.4.2 Topographic surveys
Surveys concerned with the location and representation (by plan or map) of
the main natural and artificial features of the earth’s surface including hills,
valleys, lakes, rivers, buildings, roads, railways, power lines, etc.
The topographical and engineering surveyor is often one and the same person.
The Topographical and Engineering Surveyor
Source: SAGI website - www.sagi.co.za
In order to create accurate maps for design and construction, the
Topographical and Engineering Surveyor measures and records natural and
built features on the earth's surface. He/ she also performs the surveys
necessary to control, set out and monitor the construction of buildings, roads,
bridges, dams. The work is varied and often requires high responsibility.
The topographical surveyor prepares maps necessary for all physical planning and
development. Before any work can begin, the sites for dams, bridges, canals, roads, airports,
agricultural projects and other projects must first be surveyed.
The engineering surveyor compiles a topographical map before an engineering project is
started. He/she sets out the proposed site and monitors progress of the project to ensure that
it remains within the limits as surveyed and set out by him/her.
Topographical and engineering surveyors use the trigonometrical beacons, which were
erected by the Chief Directorate of Surveys and Mapping throughout the country, as points of
reference. They make use of angular and distance measurements when recording, making
calculations and setting out construction sites.
The topographical surveyor also collects information on the names of places and annotates
information on aerial photographs by taking notes of the topographical features of the area.
He/she can prepare detailed topographical maps with the use of modern photogrammetrical
equipment.
Working conditions of topographical and engineering surveyors vary between fieldwork and
office work. They sometimes camp out in the field for days. Examination of survey records,
diagrams and plans, the work with photogrammetrical equipment, draughting and calculations
are usually done in the office.
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Topographical and engineering surveyors need good intellectual ability and must be objective
scientific observers. Mathematical aptitude, especially in Trigonometry is a necessity because
of the application in the work. Computer literacy is essential. They must be able to work on
their own or with people.
Training in South Africa is over 3 years at a University of Technology. Any
person who is registered with a University of Technology for the National
Diploma in Surveying, may register with the Council for Professional and
Technical Surveyors (PLATO) as an Engineering Survey Technician in
Training.
Upon completion of the academic qualification and the writing of a Law
Examination, a person may register with PLATO as an Engineering Survey
Technician.
After a further period of prescribed work and the writing of a Law Examination,
a person may register with PLATO as an Engineering Surveyor.
After completing a Bachelor of Technology Degree in Surveying at a
University of Technology, and after a further period of prescribed work and the
writing of a Law Examination, a person may register with PLATO as a
Professional Engineering Surveyor.
For further information, visit
www.plato.org.za
1.4.3 Cadastral surveys
Surveys concerned with the measurement of land for the preparation of plans
and diagrams, drawn to scale, showing and defining legal property boundaries
in order that ownership may be registered in the Deeds Office. Cadastral surveying
therefore involves the survey of land (townships or farms) and buildings (sectional title) for the purpose
of delimiting property boundaries and/or rights to that property.
In most countries, South Africa inclusive, these surveys are carried out by
registered Professional Land Surveyors.
The Professional Land Surveyor
Source: SAGI website - www.sagi.co.za
Professional Land Surveyors are measurement specialists and are legally
responsible for all surveys involving boundaries and land title. They are also
involved in planning, project development and professional consultation.
A Professional Land Surveyor works partly in the field. He/she uses advanced
technology and computers. He/she often works on his/her own and is
therefore dependent on his/her own judgement and decisions. The
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Professional Land Surveyor also works with the public and it is therefore
important for him/her to be able to maintain good human relations.
By law it is the exclusive function of a Professional Land Surveyor to do cadastral surveying.
Training in South Africa is over 4 years at either the University of Cape Town or the University
of KwaZulu-Natal. After completion of the formal education, land surveyors commence inservice training of 270 days. This entails 135 days of cadastral surveying and 135 days of
non-cadastral surveying, under the supervision of a registered Professional Land Surveyor
who will be responsible for the practical training. The final 30 days of this practical training are
spent at one of the regional Surveyor-General's offices where they prepare for the
examination in survey law and the surveying test. After a candidate has passed the exam and
the surveying test, he/ she can register as a Professional Land Surveyor with the SA Council
for Professional and Technical Surveyors (PLATO).
NOTE: A person who has obtained the National Diploma in Surveying and/
or the Bachelor of Technology Degree in Surveying at a University of
Technology, and who is registered with PLATO as an Engineering
Surveyor or a Professional Engineering Surveyor (see section on the
Topographical and Engineering Surveyor above), may perform, by
law, cadastral surveys under the supervision of a Professional Land
Surveyor.
1.4.4 Mine surveys
Determining the extent of existing mine works in order to keep the relationship
between surface and underground work correct, and to determine the way
mining should proceed so to reach some required position
1.4.5 Hydrographic surveys
The determination of extent of water areas, volumes, rate of flow and form
and characteristics of underwater surfaces, in connection with rivers,
harbours, lakes and along cost lines. Civil engineering works connected with
water supply, water power, irrigation, sewerage disposal, flood control, viaduct
and river works generally, also involve the practice of hydrographic surveying.
Hydrographic surveying also includes the determination of mean sea level.
1.5
Spatial Data
Spatial data, in general, refers to data associated with a location on some reference
coordinate system. The kind of spatial data used in surveying/ geomatics is
associated with location on the earth surface. Hence it is also referred to as geospatial or geo-information. This is the form of data used to represent geographic
objects on paper maps or digital spatial databases.
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1.5.1 Spatial data collection
At this phase spatial data is observed and recorded using a wide variety of
surveying instruments. The major quantities measured in the field at this
phase of data acquisition are angles and distances (horizontal and vertical).
These quantities may then be used to compute co-ordinates that used to
define the location of spatial objects.
1.5.2 Equipment for spatial data collection
Most of the equipment used in surveying/ geomatics is designed to measure
angles and distances in isolation or in combination. Each instrument is
designed to undertake a particular job and to a specific accuracy and its role
towards provision of accurate spatial data determines how much it should
cost. The cost of the instrument varies; the general rule though is "the higher
the degree of accuracy, the higher the cost.
In surveying/ geomatics, engineering surveying inclusive, instruments are
used in two opposite ways;
•
Mapping: instruments are used to construct a model of the real world by
taking measurements “in the field”. Such is the case during site surveying
for example
•
Setting out: instruments are used in the constructing of a “planned”
project as it exists in a model, into the field. Such is the case in the
setting out of “civil works” such as roads, dam walls, etc.
Some instruments can only be used for mapping and not for setting out and
vice versa. Many common instruments, however, can be used for both
mapping and setting out
Most of the instruments used in surveying/ geomatics have evolved through
different phases of human civilisation. Over time most of them have been
improved upon and therefore only modern instruments will be covered in this
course.
•
Quantities Measured: The major quantities measured in surveying/
geomatics are distances and angles. From these measured quantities
coordinates depicting locations of geographic features can be
determined.
•
Units of measurements: Measurement units used in surveying/
geomatics are distinguished between distance and angular units.
Distance Units
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The standard unit for distance measurements is an international metre (m).
Where distance measurements are recorded in multiple units such as
kilometres (km) or sub units of a metre such as decimetres (dm), centimetres
(cm) and millimetres (mm), the following conversion factors must be used.
1km = 1000m
1dm = 0.1m
1cm = 0.01m
1mm = 0.001m
Accordingly, areas are recorded in square metres (m2). In cadastral surveying,
however, there may be a requirement that the area be recorded in Hectares
(ha) when it exceeds 10000 square metres. The following conversion factors
therefore can be used:
1ha = 10000 m2
In the case of South Africa, it is important to note that the Dutch and English
settlers used different units for a foot, in the areas which were under their
control. When making conversions from very old units, the following
conversion factors must be used:
Equivalents of Units (Act No 18 of 2006):
Unit (Length) Equivalent
Foot
Foot (Cape)
Foot (Geodetic Cape)
Foot (South African Geodetic)
Inch
Mile
Nautical Mile (International)
Rood (Cape)
Rood (Geodetic Cape)
Yard
0.304 8 metre
0.314 858 1 metre
0.314 855 575 16 metre
0.304 797 265 4 metre
0.025 4 metre
1609.344 metre
1852 metre
3.778 297 2 metre
3.778 266 9 metre
0.914 4 metre
Unit (Area) Equivalent
Acre
Are
Hectare
Morgen
Square foot
Square inch
Square mile
Square yard
4 046.86 square metre
100 square metre
10 000 square metre
8 565.32 square metre
0.092 903 04 square metre
0.000 645 16 square metre
2 589 988 square metre
0.836 127 36 square metre
Angular units
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There are various units of measure for angles in use. These include the
following:
•
Degree (), Minute (’), and Second (”)
This system of measurements is also referred to as the sexagesimal
system of angular units. It is the most commonly used system for angular
measuring instruments and for calculations in engineering and surveying,
in English speaking countries.
In this system 360 degrees make up a circle, 60 minutes make a degree and 60
seconds make a minute. Angles in this measure are written for example as 1336’12” or
13.36.12. 13 degrees, 36 minutes, 12 seconds.
•
Decimal degrees.
Decimal degrees are the decimal equivalent of degrees, minutes and
seconds. Most scientific calculators calculate trigonometric functions of
angles expressed as decimal degrees, and similarly for displaying angles
calculated from inverse functions like ArcTan. Usually calculators have a
function for converting angles to and from decimal form. In this system
►
360 = 360
►
1’ = 1
60
►
1” =
= 0.017
1
= 1 = 0.0003
3600
60 x 60
Therefore 1336’12” = 13 + 0.6 + 0.0033= 13.6033
•
Grads or Gons
This system is also referred to as the centesimal system of angular units.
These are decimal angles, 400 grads is equal to a circle and therefore
equal to 360. Some European surveying instruments are based on grad
measure. Advanced scientific calculators can be set to work in Grads.
For example on some calculators you can set MODES, then select GR,
then key in angles which the calculator will assume to be in Grads units.
•
Radians
This is the natural unit for angles. An arc of one unit in a circle of 1 unit
radius subtends 1 radian. Generally computer programs expect angles to
be in radians. Whenever an angle appears in a mathematical formula (as
opposed to a trig. function of an angle) one can assume that the angle
should be in radians. It is often necessary to convert an angle from
radians to arc degrees, minutes or seconds. The symbol  (rho) is used
for the conversion factor:
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360
= 57.295 780
2
360  60  60
“=number of seconds in a radian=
= 206 265.806 25
2
=number of degrees in a radian=
1.5.3 Spatial Data Processing, Manipulation and Analysis
Once data have been collected, the resultant measurements need to be
processed, manipulated and analysed in order to obtain the required results.
The resources required for these may vary slightly depending on the mode of
data collection. The major resource for modern data processing, however, is
a computer.
The various methods of processing data, and the instruments used to collect
such data will be covered in later chapters in this course.
1.6
Maps
1.6.1 Map usefulness
A map is a two dimensional description of a specific area of land. Maps
describe in a visual or graphic format certain key features of the territory being
examined. For example, a road map shows many of the important roads, and
how they interconnect. A road map might also show the towns and cities, as
well as a few of the more popular destinations within a specified area.
Maps have been used by humans for hundreds of years. As technology has
improved, so have the quality and accuracy of maps. Ancient maps were
usually drawn by explorers. It was impossible at that time for anyone to leave
the Earth, and look down at the huge continents below. All they could do was
walk around the different land formations, and then do their best to draw what
they thought the land probably looked like.
In the modern world our ability to view and map the Earth is much improved.
However, even today, it is impossible to draw a flat map that is 100%
accurate. This is due to the impossibility of recreating the surface of a round
planet on a flat map. The smaller an area that a map represents, the more
accurate that map will be.
A map which depicts a small territory is referred to as a large scale map. This
is because the area of land being represented by the map has been scaled
down less, or in other words, the scale is larger. A large scale map only
shows a small area, but it shows it in great detail. A map depicting a large
area, such as an entire country is considered a small scale map. In order to
show the entire country the map must be scaled down until it is much smaller.
A small scale map shows more territory, but it is less detailed.
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Globes have a very important role in helping us understand the shape of our
home planet. Because globes are the same shape as the Earth, it is possible
to create very accurate globes. We do not have to worry about the distortion
that occurs when round continents are flattened out so that they fit on flat
maps.
Despite their usefulness, globes have many disadvantages as well. The
biggest of these disadvantages is due to the fact that globes represent the
entire Earth. Because of this, they will always be small scale representations
of the Earth. This means that it is difficult to show very much detail on a globe.
Only the largest objects can be depicted, such as the largest lakes and rivers.
Smaller lakes, streams, and hills simply cannot be shown. Another
disadvantage of a globe is that their awkward size and shape makes them
difficult to carry around. Globes are an important tool that students should
study. However, flat maps are much more practical for use in everyday
situations.
1.6.2 Map Projections
Over the centuries many different ways of representing the round Earth on flat
paper have been developed. Each of these methods is referred to as a map
projection.
A cylindrical projection map is the most common type of map. Areas close to
the equator have very little distortion. However, the closer to the poles that
one travels the more distorted that the map becomes, e.g. Greenland appears
to be many times larger than it really is.
A map which portrays shape accurately is called a conformal map. Conformal
maps are useful in that they help us understand the true shape of the items on
the map. However, these maps have many drawbacks. A conformal map
tends to get quite distorted, especially towards the top and bottom of the map.
This creates problems with scale. The scale may be accurate near the
equator, but the further one travels form the equator, the less accurate the
scale becomes.
1.6.3 The Transverse Mercator (Gauss Conformal) Projection
The familiar Mercator projection used on so many world maps is a cylindrical
projection, meaning the globe is encircled by an imaginary cylinder touching
at the equator, and the earth is projected onto the cylinder.
The Mercator projection is a conformal projection, meaning that angles and small shapes on
the globe project as the same angles or shapes on the map. The price paid by all conformal
projections is great variation in scale away from the central portions of the map. Greenland on
a Mercator map looks as big as South America, though it has actually only 1/8 the area.
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MAP OF THE WORLD: MERCATOR PROJECTION
There is no reason the cylinder has to touch at the equator, though. An
alternative choice would be a Mercator projection whose cylinder touches the
earth along a meridian of longitude. Such a projection is called a Transverse
Mercator projection.
MERCATOR PROJECTION (LEFT)
TRANSVERSE MERCATOR PROJECTION (RIGHT)
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Chapter 2
THE SOUTH AFRICAN COORDINATE SYSTEM,
AND BASIC COORDINATE CALCULATIONS
2.1
Coordinates
In survey work measured slope distances are transformed into their horizontal
and vertical components. The horizontal distances are used when plotting a
plan, and these horizontal distances, together with "directions" are
transformed into two further components called COORDINATES.
The use of the trigonometrical functions, "sine" and "cosine" will, together with
the horizontal distance, give the component distances along the axes.
The position of any point can be fixed relative to these axes, by measuring the
point's perpendicular distance from each of the axes, the intersection of which
is called the ORIGIN.
In mapping and survey work in South Africa, these coordinates are referred to
as the Y and X coordinates of the point.
It is usual South African practice for Y, the ORDINATE, to represent the
measurement from East to West, and X, the ABSCISSA, to represent the
measurement from North to South.
It is a convention that Y be written before X, thus:
- 22 901,14Y + 3 316 209,06X
NOTE:
2.2
A coordinate without a sign is meaningless and, on no
account, should the sign be omitted.
The General Triangulation of a Country
Triangulation is the process of determining the positions of points by means of
triangles. This is the meaning of the word in its narrowest sense but,
according to generally accepted usage, it has a far wider application; the
determination of positions by the intersection of observed direction rays.
2.2.1 Basic Procedure - The progressive steps in establishing a
triangulation system are:
2.2.1.1
The selection of circuits - the country is divided into sections
or circuits as they are called.
Approximate squares of about 500-800 km sides are selected, in
such a way as to offer a suitable site near each corner, where a
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base line may be measured accurately, and so that a geodetic
chain of triangles may be extended between these base lines.
2.2.1.2
Base line measurement - the triangulation system is based
upon measured lines, known as BASE LINES, whose terminal
points are fixed astronomically. The length of the base is
extended by a process of triangulation, known as "base line
extension", after which the further steps enumerated below are
carried out, also by triangulation process.
The base lines must be measured very accurately, and the
terminal points marked with symmetrical concrete beacons,
accurately centred over the terminal points. Base lines vary in
length from about 8-30km, depending on:
i.
ii.
the suitability of the site for measuring;
the length required to give well-conditioned extension
triangles from the base line. (By "well-conditioned"
triangle, is meant a triangle which is nearly equilateral as
possible).
The distance apart of the base lines is usually about 20-40 times
the length of the bases, but may vary between even wider limits.
2.2.1.3
Geodetic Chain - must be composed of very well-conditioned
triangles, where sides are as long as possible. In South Africa
the sides are, on the average, about 60km long, but they vary
from 30km to 190km, depending on the terrain.
2.2.1.4
Primary Triangulation - the areas inside the Geodetic Chains
are covered with a network of triangulation, the beacons and
observations being of the same type and accuracy as before.
The side of a primary triangle is in the neighbourhood of 3060km, depending upon the topography.
2.2.1.5
Secondary Triangulation - consists of filling in each primary
triangle with triangles of about 12-18km sides (see Fig. ). The
beacons used are the same type of concrete beacon but, in
place of the quadruped structure, a metal signal or vane is set
symmetrically onto the top of the beacon.
The signal may be removed to allow a theodolite to be set up
over the centre of the beacon.
2.2.1.6
2.2.1.7
Tertiary Triangulation - consists of breaking down the
secondary triangulation to triangles about 3-6km sides. These
beacons are usually 38cm diameter and 1,2m high, surmounted
by a metal signal 1,2m high.
Reference marks in townships - this is the final breakdown of
coordinated points and consists of marked traverse stations,
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usually under small circular metal covers, set into the road
surface, in townships and built up areas. From these the
coordinates of erven and lots may be determined readily, and
the positions of their beacons referenced. This part of the
programme for South Africa's coordinate system is far from
completed.
2.3
The South African Coordinate System
The South African coordinate system is based on the "GAUSS
CONFORM PROJECTION". This system consists of belts running
NORTH and SOUTH, two degrees of longitude wide, with the central
meridian being the odd meridian.
▪
The Y coordinate is measured positive to the west of the origin and negative
to the east. The X coordinate value increases from zero at the equator in a
southern direction and, therefore, always positive in South Africa.
.
In the northern hemisphere and, in some systems, in the southern
hemisphere, these conventions are reversed. In survey work angles
are always measured in a clockwise direction.
▪
▪
The co-ordinate grid is superimposed onto the geographical graticule, the
zero Y grid line being collinear with the central meridian. The meridians
converge as we move from the equator towards the pole, and will be
symmetrical about the central meridian in each belt.
The co-ordinates of a point on the boundary meridian between two
adjacent systems, i.e., the even meridian, will have the same X value, and
the Y will have the same numerical value, but will have opposite signs in
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▪
▪
▪
SURVEYING: THEORY AND PRACTICAL
the two systems. If a survey falls on or near the boundary meridian, it is
usual to calculate the entire survey in the belt in which the greater portion of
the area falls. For this purpose the co-ordinates of trig beacons falling within
an overlap area of 15 minutes of longitude on either side of the boundary
meridian, are given on both co-ordinate systems.
It should be noted that the X value always gives the true distance of the point,
measured along the central meridian, from the equator. The Y co-ordinate
gives the distance east or west of the central meridian and it is here that
scale distortion is present. On the map these lines are longer than their actual
length on the earth's surface.
Bearings or Azimuths are true geographical angles of direction and are
related to true north or true south.
Hence the bearing of a line and its direction on the co-ordinate system will
only coincide when the point of measurement is situated on the central
meridian. The word "bearing" should thus not be used when dealing with a
rectangular co-ordinate system.
For short lines, their directions on the map are virtually the same as their
bearings on the spheriod, and as a result, the shapes of small figures are
similar to their shapes on the spheroid although, due to scale distortion,
their areas are increased proportionally to their distances away from the
central meridian.
2.3.1 The Grid and Graticule
In most survey work coordinates are used, and before plotting them, a
rectangular "coordinate grid" is constructed.
A GRID is the representation on a map of a system of equally spaced
straight lines, parallel to the Y and X axes of the coordinate system, the
exact distance of each line from its parent axis being known. The grid
thus consists of a system of squares of known dimensions.
Knowing the value (distance from the zero axis) of the grid lines, the
position of a point of known coordinates can be plotted by scaling the
correct distance from two nearby grid lines.
A GRATICULE is the representation of the lines of latitude and
longitude on a map.
2.3.2 Angles of Direction
The direction, angle of direction, or direction angle of any line on any
coordinate system is the angle, measured in a clockwise sense,
between zero direction of the coordinate system and the line.
The zero direction of the South African system is the direction in which
the X coordinate increases positively, whilst the Y coordinate remains
unchanged, ie. true geographical south at the central meridian of each
belt.
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A direction is always referred to the zero direction of the coordinate
system, and not to any other line, such as true magnetic north. On
small local survey directions are sometimes referred to any arbitrary
line and, in such cases, they are called "Assumed Directions". A line
pivoted at the origin and rotated clockwise starting from the positive
direction of the X axis, will sweep out lines of direction starting form 0
to 360.
The first quadrant contains direction angles from 0 to 90.
The second quadrant contains direction angles from 90 to 180.
The third quadrant contains direction angles from 180 to 270.
The fourth quadrant contains direction angles from 270 to 360.
2.3.3 Orientation
Orientation is the rotation of an angle so as to bring the initial line to the
zero direction (in South Africa this is south).
The coordinate values of the control network over the Republic are
supplied, by the Trigonometrical Survey Office, in metres. The heights
of the Trigonometrical Stations (trig stations) above mean sea level are
given to the tops of all trig beacons in metres, unless otherwise stated
in the trig lists.
Mean Sea Level is determined by taking tide observations at various
points around the coast, over a continuous period for many years.
From these observations, a mean height of the sea is determined
accurately.
The most accurate method of surveying heights is by precise levelling,
and a network of precise levels connected to the tide gauges has been
carried out. The heights of the trig beacons are connected to the
precise benchmarks. These elevations are determined by observing
vertical angles and adjusting the heights to conform with those of the
precise levelling network where possible.
This operation is known as trig levelling. In this way, heights can be
carried over the whole country, although many of the trig stations are
great distances from the nearest line of precise levels.
2.4
Calculations (Coordinate) - South African Coordinate System
The map projection used is the Transverse Mercator Projection (also known
as the Gauss Conformal). This is a cylindrical projection, and has the
advantage that directions remain true for short lengths as you move from the
central meridian.
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2.4.1 Reference Systems
GRATICULE is a reference system using lines of longitude and
latitude, ie. it is a geographical system.
GRID - the Transverse Mercator projection has a rectangular grid. The
N-S line agrees at the central meridian only.
DISTANCES - are measured on the graticule and must be corrected to
sea level and allowance made for scale distortion.
DIRECTIONS - note that sometimes the term "direction" and "bearing"
are used interchangeably; in South Africa we use "bearing" related to
geographical systems and "direction" related to the grid system.
TRUE NORTH (T.N.) - the true north at a point is the direction of the
North Pole from that point.
The meridian through any point is a true North-South line. We have
shown that meridians converge towards the poles. Therefore, true
North directions also converge towards the Poles. The T.N. at two
points will only be parallel lines if they are on the Equator or on the
same meridian.
It is apparent, therefore, that the T.N. line is not a good direction on
which to base the directions used in a survey, because it is not a
parallel direction from all points in the survey.
GRID NORTH (G.N.) - the central meridian of each system is a T.N.
(T.S.) line, but all grid lines are made parallel to or at right angles to
this line and, for survey purposes, are oriented from the Grid North
(South) as it is called, ie. a line parallel to the central meridian.
The angle of direction or the direction of a line is always referred to as
G.S. in South Africa. Although these directions are fairly commonly
referred to as "bearings", this is a misnomer, as true geographical
bearings refer to T.N.
MERIDIAN CONVERGENCE - G.N. coincides with T.N. only at the
central meridian of each zone, e.g. as the 31 meridian passes through
Durban, T.N. and G.N. coincide here.
The difference between G.N. and T.N. (G.S. and T.S.) is of opposite
sign on either side of the central meridian. By drawing a figure, the sign
of this meridian convergence can easily be determined. A maximum
value for meridian convergence in South Africa is about 30 minutes of
arc.
MAGNETIC NORTH (M.N.) - is the direction in which a freely
suspended compass needle points.
18
SUTP101
SURVEYING: THEORY AND PRACTICAL
2.4.2 The Terms used in Coordinate Calculations
a.
b.
c.
d.
e.
f.
g.
Direction is the angle measured from a reference object in a
clockwise sense.
Oriented direction (direction angle) is the angle measured from
grid south. At the central meridian this coincides with
geographical or true south.
Functional angle - this is to simplify calculations when using
logarithmic or natural tables. The function angle is the direction
minus the preceding right angle, e.g. the functional angle of 123
is 33 and 327 is 47.
Orientation is the measure of the rotation of an angle so as to
bring the initial line to zero direction, i.e. grid south.
Zenith - an imaginary point vertically above the instrument.
Zenith distance is the angle between the zenith and the point
sighted. It is the vertical angle read on most modern theodolites.
Standard methods of calculation. The civil engineering industry
follows the standard method laid down by the Surveyor General
for cadastral surveys.
First quadrant of coordinate system:
+Y
YB
Y = Y-difference
YA
90
A
XA

S
X
Y
B
XB
+X
19
0
SUTP101
SURVEYING: THEORY AND PRACTICAL
2.4.3 The Polar on Electronic Calculator
Dir AB
A
Dist AB
B
ΔY = Dist AB x Sin of Direction AB
YB =
YA + (±ΔY)
YA
XA
ΔY
ΔX
YB
XB
ΔX = Dist AB x Cos of Direction AB
XB = XA + (±ΔX)
Check the polar by doing a join separately in its standard form.
NOTE: Polar calculations do not have to be presented in the above format. The
following format is acceptable and recommended:
A-B
x AB
Dist AB
B
YB
XB
2.4.4 The Join on Electronic Calculator
A
YA
XA
Dir AB
B
YB
XB
Dist AB
ΔY = YB -YA
ΔX = XB - XA
Check B
YB √
XB √
NOTE: Join calculations do not have to be presented in the above format. The
following format is acceptable:
A–B
x AB
Checked by Polar √
Dist AB
Step 1:
Find ΔY and ΔX by subtraction
Step 2:
Find x AB:
a.
 Y 
Always obtain a value for Tan −1 

 X 
b.
Place in proper quadrant by inspecting the signs of ΔY and ΔX.
Add 0º if the signs are
+
+
(See diagram of first quadrant on
previous page)
Direction AB = x AB = α
20
SUTP101
SURVEYING: THEORY AND PRACTICAL
-X
B
180
XB

1
S
X
Y
+Y
90
YB
Y = Y-difference
See diagram of the second quadrant above.
 Y 
α 1 = Tan −1 

 X 
 ΔY 
α = 90 o −  1 = 90 o − Tan −1 

 ΔX 
 +Y 
 + 90º + 90º
x AB = α + 90 o = Tan −1 
 −X 
 Add 180 º if the signs are
+
−
21

A
YA
XA
SUTP101
-X
SURVEYING: THEORY AND PRACTICAL
180
Y
XB
B
X
S

XA
A
YA
270
Y = Y-difference
See diagram of the third quadrant above.
Direction AB = x AB = α + 180 º
 Add 180 º if the signs are
−
−
PTO FOR FOURTH QUADRANT
22
YB
-Y
SUTP101
SURVEYING: THEORY AND PRACTICAL
YA
XB
Y = Y-difference
-Y
YB
270
Y
A

X
S

1
XA
B
+X
0
See diagram of the fourth quadrant above.
 Y 
α 1 = Tan −1 

 X 
 ΔY 
α = 90 o −  1 = 90 o − Tan −1 

 ΔX 
 −Y 
 + 90º + 270º
x AB = α + 270 o = Tan −1 
 +X 
 Add 360 º if the signs are
−
+
Step 3: Find the distance AB:
ΔY
or
sin x AB
AB
=
or AB =
ΔY 2 + ΔX 2
ΔX
cos x AB
AB =
(best to use)
23
SUTP101
SURVEYING: THEORY AND PRACTICAL
EXERCISES
1.
Calculate checked coordinates for point B from the data below. Your answer should
be, within a few mm’s, the same for both polars.
A
Dir AB
+ 571,436
103º55'26"
-144,770
Dist AB
362,498m
D
Dir DB
+842,342
11º45'38"
-620,777
Dist DB
397,114m
Solution:
NOTE: FOR JOIN AND POLAR CALCULATIONS ONLY THE TEXT PRINTED IN
BOLD ITALICS BELOW NEEDS TO BE SHOW IN TESTS AND THE EXAM
1.
POLAR
=====
From
====
A
A
To
==
Direction
=========
Distance
========
B
103.55.26
362.498
From
====
D
D
To
==
Direction
=========
Distance
========
B
11.45.38
397.114
Y
=
+571.436
+923.282
X
=
-144.770
-231.999
Y
=
+842.342
+923.283
X
=
-620.777
-231.999
2.
Calculate the reduced distance and the oriented direction from A to E. Your
answer must be fully checked.
A
E
+ 4932,566 + 7892,023
+ 3961,301 + 8371,822
Solution:
2.
From
====
A
A
JOIN
====
To
==
Direction
=========
Distance
========
E
296.17.21
1083.311
Y
=
+4932.566
+3961.301
X
=
+7892.023
+8371.822
CHECKED BY POLAR
3.
Give, in point form, a brief description of the South African Coordinate
System. Support your answer with suitable sketches.
4.
Coordinates:
y
x
24
SUTP101
SURVEYING: THEORY AND PRACTICAL
A
B
C
D
+ 324,413
+ 553,813
+ 746,317
+1033,285
+ 468,229
+ 869,194
+ 399,115
+ 819,726
4.1
Calculate joins AB, AC and AD.
5.
Coordinates:
y
A
C
E
x
+ 296,443
+ 774,494
+ 155,202
y
+ 486,326
+ 548,323
+ 875,042
5.1
Calculate joins AB, AC, AD and AE.
5.2
Direction
BF
=
B
D
x
+ 566,923
+1044,349
209 : 52 : 33 Distance
BF
+ 842,899
+ 709,904
=
312,522m
Calculate unchecked coordinates for F.
6.
Calculate the reduced distance and the oriented direction from J TO K:
J
K
- 5247.331
- 7244.316
+8422.725
+8111,724
Your answer must be fully checked.
7.
Calculate checked coordinates for point L from the data below. Your answer
should be, within a few mm’s, the same for both polars.
R
Dir RL
+ 240,429
+8012,395
182º 06'12" Dist RL
335,705m
Z
Dir ZL
+229,481
+7688,324
186º 52'01" Dist ZL
11,491m
8.
Calculate the reduced distance and the oriented direction from DRUM to POLE:
DRUM
POLE
+ 2744,217 + 5621,498
+ 1821,301 +8266,613
25
SUTP101
9.
SURVEYING: THEORY AND PRACTICAL
Calculate checked coordinates for point SHACK from the data below. Your
answer should be, within a few mm’s, the same for both polars.
BORDER
+ 261,324
Dir BORDER – SHACK
-413,668
128º 26'28" Dist BORDER – SHACK
943,095m
DUKE
+927,401
Dir DUKE – SHACK
-554,328
170º 44'52" Dist DUKE – SHACK
451,546m
10. Calculate the reduced distance and the oriented direction from CABLE to BON
ACC:
CABLE
BON ACC
+ 1682,47
+ 5244,68
+ 3333,10
+1919,52
11. Calculate the reduced distance and the oriented direction from CABLE to JUMI:
RES8
JUMI
+ 1827,45
+ 5526,31
+ 9273,81
+1124,69
12. Calculate the reduced distance and the oriented direction from DURR to KP:
DURR
KP
+ 5527,45
+ 2651,98
+ 1891,54
+10 026,87
13. Calculate the reduced distance and the oriented direction from PENN to IFAFA:
PENN
IFAFA
14.
+ 19 261.54 + 5261.30
+ 16 609,18 +9111,11
Calculate the reduced distance and the oriented direction from MHLOTI to
UM-R:
MHLOTI
UM-R
+ 1818,46
+ 2424,92
+ 19 999,99
+26 060,46
PTO FOR SOLUTIONS
26
SUTP101
SURVEYING: THEORY AND PRACTICAL
Solutions
4.
JOINS: ALL CHECKED BY
POLAR
=====
Y
=
+324.413
+553.813
+746.317
+1033.285
From
X
====
=
A
+468.229
A
+869.194
A
+399.115
A
+819.726
To
Direction
Distance
==
=========
========
B
29.46.28
461.949
C
99.18.12
427.527
D
63.37.31
791.233
5.
JOIN
====
Y
=
+296.443
+155.202
From
X
====
=
A
+486.326
A
+875.042
To
Direction
Distance
==
=========
========
E
340.01.53
413.581
CHECKED BY POLAR
NOTE: STUDENTS TO DO JOINS AB, AC and AD and compare answers
POLAR (unchecked)
=====
Y
=
+566.923
+411.249
6.
Page i of 285
From
X
====
=
B
+842.899
B
+571.909
To
Direction
Distance
==
=========
========
F
209.52.33
312.522
JOIN
====
SUTP101
Y
=
5247.331
7244.316
SURVEYING: THEORY AND PRACTICAL
From
X
====
=
J
+8422.725
J
+8111.724
To
Direction
Distance
==
=========
========
-
K
261.08.53
2021.057
CHECKED BY POLAR
7.
POLAR
=====
Y
=
+240.429
+228.108
Y
=
+229.481
+228.107
8.
Y
=
+2744.217
+1821.301
From
X
====
=
R
+8012.395
R
+7676.916
From
X
====
=
Z
+7688.324
Z
+7676.915
To
Direction
Distance
==
=========
========
L
182.06.12
335.705
To
Direction
Distance
==
=========
========
L
186.52.01
11.491
JOIN
====
From
X
====
=
DRUM
+5621.498
DRUM
+8266.613
To
Direction
Distance
==
=========
========
POLE
340.45.56
2801.501
CHECKED BY POLAR
POLAR
=====
9.
Y
=
From
X
====
=
To
Direction
Distance
==
=========
========
ii
-
SUTP101
+261.324
+1000.001
Y
=
+927.401
+1000.001
SURVEYING: THEORY AND PRACTICAL
BORDER
-413.668
BORDER
-1000.000
From
X
====
=
DUKE
-554.328
DUKE
-999.999
SHACK
128.2628
943.095
To
Direction
Distance
==
=========
========
SHACK
170.4452
451.546
JOINS: ALL CHECKED BY
10 - 14
POLAR
=====
Y
=
+1682.470
+5244.680
+1827.450
+5526.310
+5527.450
+2651.980
+19261.540
+16609.180
+1818.460
+2424.92
From
X
====
=
CABLE
+3333.100
CABLE
+1919.520
RES8
+9273.810
RES8
+1124.690
DURR
+1891.540
DURR
+10026.870
PENN
+5261.300
PENN
+9111.110
MHLOTI
+19999.990
MHLOTI
+26060.46
To
Direction
Distance
==
=========
========
BONACC
111.38.40
3832.434
JUMI
155.35.13
8949.286
KP
340.32.02
8628.553
IFAFA
325.26.05
4675.046
UM-R
05.42.52
6090.738
iii
SUTP101
SURVEYING: THEORY AND PRACTICAL
Chapter 3
SPIRIT (DIFFERENTIAL) LEVELLING
3.1 Useful Terminology
•
Datum: The elevation of a point has to be expressed relative to a certain
DATUM. This DATUM can be any arbitrary point or surface (often referred to
as a LOCAL DATUM). The most commonly used datum, however, is MEAN
SEA LEVEL (MSL).
When a Surveyor takes levels on a building site for the purpose of designing a
building (eg. a house), any permanent convenient point in the vicinity, eg. a
manhole in the road, can be given an arbitrary height. When heights are set
out from the building plan at the construction stage, the same point or points
can be used.
•
The Direction of Gravity is the direction assumed by a free-hanging or
suspended plumb-bob. A Vertical Plane is defined as a plane that contains
the plumb line.
•
A Level Line or Level Surface is a line or surface on which all points are of
equal elevation, and is perpendicular to the direction of gravity at that point.
•
A Horizontal Line is a straight line, and is tangential to the level line. A
levelling instrument that is in perfect adjustment provides the user with a
horizontal line of sight, also known as the line of collimation.
•
The elevation of a point relative to the datum in use is called the Reduced
Level of the point.
•
A Bench Mark is a point installed in such a way that movement of the point in
a vertical or horizontal direction is negligible. They can be brass studs set in
concrete, bridges or other permanent structure. Their Reduced Levels are
accurately determined.
iv
SUTP101
3.2
SURVEYING: THEORY AND PRACTICAL
•
A Backsight is the first reading taken onto a staff after the levelling instrument
(level) is set up.
•
A Foresight is the last reading taken onto a staff before the level is moved
away from the particular instrument position.
•
An Intermediate Sight is taken after the backsight and before the foresight of
that particular instrument position.
The Level basically consists of a telescope with a bubble tube attached to it.
The telescope consists of an objective lens, focusing lens, focusing screw,
diaphragm and eyepiece. The objective lens is the large lens at the front end of the
telescope. The eyepiece is located at the other end. The diaphragm consists of plane
glass upon which the cross-hairs and stadia hairs are engraved.
THE IMAGINARY LINE JOINING THE CENTRE OF THE CROSS HAIRS AND THE
OPTICAL CENTRE OF THE OBJECTIVE LENS IS CALLED THE LINE OF
COLLIMATION OR LINE OF SIGHT.
The telescope is mounted on a TRIBRACH, which is usually equipped with
three footscrews. The tribrach can be mounted on a tripod.
3.2.1 Types of Levels
a.
The Dumpy Level: called "dumpy" because of its shorter and thicker telescope
than the other levels at the time it first made its appearance. On many building
sites any survey instrument is often incorrectly referred to as a "dumpy".
The main feature of the dumpy is that the telescope is rigidly attached to a
vertical axis. For accurate work, therefore, the bubble tube axis must be at
right angles to the vertical axis, and subsequently the line of collimation must
be parallel to the bubble tube axis.
b.
The Tilting Level: The major difference between the tilting level and the dumpy
is that the telescope of the tilting level is not rigidly attached to a vertical axis,
and can be tilted in the vertical plane by a small amount. The bubble axis and
the line of collimation should be parallel and, when the line of sight is
horizontal, the vertical axis need not be truly vertical.
Once the instrument is levelled approximately, by means of the circular
bubble, the telescope is levelled exactly, by means of the tilting screw, for
each different reading.
c.
The Automatic Level. In both the dumpy and tilting levels a truly horizontal line
of sight could only be obtained if the user set the equipment up correctly. The
5
SUTP101
SURVEYING: THEORY AND PRACTICAL
automatic level, on the other hand, only requires approximate levelling of a
circular plate bubble. The horizontal line of sight is provided by a system of
compensators, which will provide a horizontal line, even when the optical axis
of the telescope is not exactly horizontal.
d.
The Laser Level: Two general classes of laser levels are in use in surveying
today: Single Beam Lasers and Rotating Beam Lasers. Lasers can be
projected any direction in the vertical or horizontal planes. Lasers are very
useful for the vertical control or pipeline, tunnel, parking lot and other types of
construction.
3.3
Taking a Reading
3.3.1 The Levelling Staff is usually made of wood or of aluminium alloy. They are
available in a variety of lengths up to 6 metres. Some of the shorter staves are
in one piece, whilst others are of the telescopic type.
The most common graduation pattern used on the face of the staff is the socalled 'E' pattern.
A circular hand bubble can be used to hold the staff vertical during
observations, by positioning it against a corner of the staff. Some staves are
equipped with bubbles which are attached to the back of the staff.
For very accurate or precise work, special INVAR staves are used. (Invar is an
alloy with a very low coefficient of expansion). A parallel plate micrometer is
attached to the level for precise readings, often to 2 decimal places of a
millimetre.
6
SUTP101
SURVEYING: THEORY AND PRACTICAL
3.3.2 Change Points or Change Plates usually consist of a triangular metal plate
with three sharp feet with a rounded raised centre on the upper side.
Change points are usually used when the surface on which the staff is to be
placed is not firm or hard but sloping. When the staff is turned an error occurs.
3.3.3 The Levelling Procedure
3.3.3.1
The levelling line usually starts at a point of known elevation. This point
could have been determined from previous levelling on the same
project (contract), or it could be a Bench mark with known (published)
height.
3.3.3.2
Place the level in a convenient position near the starting point, ensuring
that a horizontal line of sight will intersect the staff. Level the
instrument. The first reading can now be taken - it is called a
BACKSIGHT.
3.3.3.3
If a reading from this instrument position to a staff at the next peg to be
levelled is impossible, the change point must be placed in a convenient
position for the forward reading before the instrument is moved. This
reading is called a FORESIGHT.
7
SUTP101
SURVEYING: THEORY AND PRACTICAL
3.3.3.4
Any reading taken, for whatever reason, after the backsight and before the
foresight, is called an INTERMEDIATE SIGHT.
3.3.3.5
After the foresight is taken, the instrument is moved to the next
instrument position and a reading is taken on the last staff position of
8
SUTP101
SURVEYING: THEORY AND PRACTICAL
the previous set up (in other words the last foresight). This reading is
again called a backsight.
3.3.3.6
Repeat this procedure until the closing (end) point is reached. This
point will also have a known elevation, and could be the same point at
which the levelling was started.
3.3.3.7
Inverted staff readings
•
•
3.3.3.8
staff held upside down with the base on the point
reading taken recorded as negative value on the levelling booking form
Two Peg Test (for checking Collimation error)
•
Set up 1: Instrument at equal distances from the staves
▪
▪
▪
place level at equal distances of 25m from the staffs
take readings 'hA1' and 'hB1'
true height difference between A and B is:
H AB1 = hA1 − hB1
9
SUTP101
SURVEYING: THEORY AND PRACTICAL
•
Set up 2: Instrument at short distance beyond staff B
▪
▪
▪
place level at a very short distance (5m) from staff B
take readings 'hA2' and 'hB2'.
height difference between A and B is given by the expression:
H AB 2 = hA2 − hB2
•
The collimation error therefore would be determined as:
=
H AB2 − H AB1 (hA2 − hB2 ) − (hA1 − hB1 )
=
D AB
D AB
Note:
▪
In set up 1, error in the instrument is eliminated due to the distances to the
staves being equal.
▪
In set up 2, the observed height difference will be different from that
observed when the instrument is positioned half way the staff positions if
the instrument has a collimation error.
▪
Therefore collimation error is eliminated if BS distance is equal or
approximately equal to FS distance
3.3.4 Booking the Readings
3.3.4.1
The first backsight is booked on the first line of the backsight column.
3.3.4.2
Book the following reading(s) on the lines corresponding with the name
of each peg in the remarks column. Intermediate sights (if any) are
booked in the Intermediate Sight column and the last reading before the
instrument is moved, is booked in the foresight column.
10
SUTP101
SURVEYING: THEORY AND PRACTICAL
3.3.4.3
The backsight from the following instrument position is booked on the
same line as the previous foresight, so that readings to the same point
are booked on the same line, but in their different columns.
3.3.4.4
The procedure is continued until the last foresight to the end point is
booked.
3.3.5 The Observation Procedure at each instrument position should entail the
following:
3.3.5.1
3.3.5.2
3.3.5.3
3.3.5.4
3.3.5.5
3.3.5.6
3.3.5.7
3.4
Set up and level the instrument
Take the reading on the staff, estimating the millimetres where the
horizontal hair intersects the staff.
Check the bubble.
Book the reading.
Check the reading on the staff.
Check the entry in the field book.
Repeat all the steps from 2.4.5.2 - 2.4.5.6 above, for each subsequent
reading at any particular instrument position.
Reducing the Observations
3.4.1 The Rise-and-Fall Method
The difference in height between two successive points is calculated by subtracting
each staff reading (except, of course, the first backsight at each set up) from the
previous staff reading.
Therefore, between points A and P1, the difference in height is 1,618 - 0,896 = 0,722.
The positive sign of the answer indicates that it is a RISE. Let us consider some of
the other observations:
P1-P2 :
P2-P3 :
2,599 - 1,413 =
1,561 - 2,799 =
+1,186 (Rise)
-1,238 (Fall)
In a field book the calculations are shown in bold italics below:
POINT
BS
IS
FS
A
1,618
P1
2,599
0,896
P2
1,561
1,413
RISE
RL
Cor
AL
FL
12,645
0
12,645
12,645
0,722
13,367
+1
13,368
1,186
14,553
+2
14,555
11
FALL
SUTP101
SURVEYING: THEORY AND PRACTICAL
P3
2,799
MH1
2,133
P4
1,555
G
1,238
1,317
13,318
13,981
+3
13,984
13,203
+3
13,206
13,441
+4
13,445
0,650
12,791
+4
12,795
2,666
12,645
0,778
0,238
1,967
7,333
+3
0,666
2,911
B
13,315
7,187
2,812
7,187
2,666
0,146
0,146
12,795
0,146
The last column (Final Level) is only completed if the line is levelled more than once.
3.4.2 The Collimation Method - in this method all elevations are obtained after first
calculating the height of the collimation line at each instrument position.
Considering the following example:
POINT
BS
A
1,618
P1
2,599
P2
1,561
IS
FS
COLL LINE
RL
Cor
AL
FL
14,263
12,645
0
12,645
12,645
0,896
15,966
13,367
+1
13,368
1,413
16,114
14,553
+2
14,555
P3
2,799
13,315
+3
13,318
MH1
2,133
13,981
+3
13,984
13,203
+3
13,206
13,441
+4
13,445
1,967
12,791
+4
12,795
7,187
12,645
P4
1,555
G
2,911
14,758
1,317
B
7,333
7,187
12,795
0,146
0,146
The last column (Final Level) is only completed if the line is levelled more than once.
3.5
Comparing the Two Methods
12
SUTP101
SURVEYING: THEORY AND PRACTICAL
a.
Rise-and-Fall Method: By adding the backsights and foresights and obtaining
the difference in the totals, an arithmetic check can be carried out with the
Rise-and-Fall method. The difference in the totals of the Rises and Falls
should be the same as the difference calculated for total Backsights and
Foresights. Since each reading was used to obtain the Rises and Falls, all the
arithmetic is checked up to this point. The two equal differences calculated up
to this point must also be equal to the difference between the reduced levels
of the start and end points.
b.
Collimation Method: In this method each reduced level is not obtained from
the previous one and, if an error was made in the reduction of an intermediate
sight, it is not carried forward and will go unnoticed.
3.6
Sources of Error in Levelling
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Mistakes in staff reading and booking
Disturbing the level
Failure to level the bubble
Incorrect focussing (parallax not eliminated)
Mistakes in reducing
Staff not being held upright
Unequal back- and foresight lengths
Telescopic staff not properly extended
Instrument not set up on firm ground
Strong wind
Heat shimmering
Sun shining on instrument
Play in tripod joints
Staff graduation errors
Staff bubble out of adjustment
EXERCISES - Two Peg Test
1. During a Two Peg Test (for checking collimation error), the difference in height
between two points was established as 1,250m. If the reading on the one staff
after moving the instrument was 1,650. Calculate what the reading on the other
staff should be for no collimation error . (Ans: 2,900)
Do calculation here:
2. A levelling instrument was set up midway between two staffs, A and B, 60 m
apart. The readings taken were 1,185 and 1,729 on A and B respectively. The
level was then moved and set up near staff A. The reading on staff A was then 1,
397. Calculate what the reading on staff B should be, if there is no collimation
error. (Ans: 1,941)
13
SUTP101
SURVEYING: THEORY AND PRACTICAL
Do calculation here:
3. A levelling instrument was set up midway between two staffs, A and B, 60 m
apart. The readings taken were 1,197 and 1,612 on A and B respectively. The
level was then moved and set up near staff A. The reading on staff A was then
1,458. Calculate what the reading on staff B should be, if there is no collimation
error. (Ans: 1,873)
Do calculation here:
EXERCISES: Reduction of levelling field book using both methods.
Question 1
POINT
BS
BM91
2,424
CP1
IS
FS
RL
Cor
AL
FL
1,665
1,617
4,213
CP3
0,702
3,115
CP4
1,419
BRIDGE
-1,828
BM92
FALL
101,913
CP2
CP5
RISE
1,010
4,881
2,739
92,726
14
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
BM91
2,424
CP1
IS
FS
RL
Cor
AL
FL
101,913
1,665
CP2
1,617
4,213
CP3
0,702
3,115
CP4
1,419
BRIDGE
-1,828
CP5
COLL LINE
1,010
4,881
BM92
2,739
92,726
Question 1 - Solutions
POINT
BS
BM91
2,424
CP1
IS
FS
1,665
RISE
FALL
0,759
RL
Cor
AL
FL
101,913
0
101,913
101,913
102,672
+2
102,674
CP2
1,617
4,213
2,548
100,124
+2
100,126
CP3
0,702
3,115
1,498
98,626
+4
98,630
0,717
97,909
+6
97,915
101,156
+6
101,162
CP4
1,419
BRIDGE
-1,828
CP5
1,010
BM92
5,753
3,247
4,881
6,709
94,447
+6
94,453
2,739
1,729
92,718
+8
92,726
13,201
101,913
14,948
4,006
14,948
13,201
-9,195
-9,195
15
-9,195
92,726
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
BM91
2,424
CP1
IS
FS
COLL LINE
RL
Cor
AL
FL
104,337
101,913
0
101,913
101,913
102,672
+2
102,674
1,665
CP2
1,617
4,213
101,741
100,124
+2
100,126
CP3
0,702
3,115
99,328
98,626
+4
98,630
CP4
1,419
97,909
+6
97,915
BRIDGE
-1,828
101,156
+6
101,162
94,447
+6
94,453
2,739
92,718
+8
92,726
92,726
14,948
101,913
AL
FL
CP5
1,010
4,881
BM92
5,753
95,457
14,948
-9,195
-9,195
Question2
POINT
BS
TSM55
1,771
P
1,315
IS
CP1
1,424
T
1,419
GATE
RL
Cor
0,899
1,106
0,912
U
V
FALL
1,619
1,880
S
RISE
242,782
Q
R
FS
1,204
0,885
1,732
1,321
2,072
242,826
16
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
TSM55
1,771
P
1,315
IS
CP1
1,424
T
1,419
Cor
AL
FL
0,899
1,106
0,912
U
V
RL
1,619
1,880
S
COLL LINE
242,782
Q
R
FS
1,204
0,885
1,732
1,321
GATE
2,072
242,826
Question 3
Note: Some field books do not make provision for the correction to be shown in a
separate column – “Cor” in the examples above. The correction is shown in the “RL”
column and entered in small print above each reduced level. Do the rest of the
examples using such a field book format.
POINT
BS
ABM 12
1,412
J1
1,725
IS
J3
1,318
ABM 16
FALL
RL
AL
FL
1,319
1,527
J5
RISE
428,719
J2
J4
FS
2,122
1,882
1,791
2,444
428,330
17
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
ABM 12
1,412
J1
1,725
IS
RL
AL
FL
1,319
1,527
J3
1,318
2,122
J5
COLL LINE
428,719
J2
J4
FS
1,882
1,791
ABM 16
2,444
428,330
Question 3 - Solutions
POINT
BS
ABM 12
1,412
J1
1,725
IS
FS
1,319
RISE
FALL
RL
AL
FL
428,719
428,719
428,719
0,093
428,812 -1
428,811
J2
1,527
0,198
429,010 -2
429,008
J3
1,318
0,209
429,219 -2
429,217
428,655 -2
428,653
428,986 -3
428,983
0,653
428,333 -3
428,330
1,217
428,719
J4
2,122
J5
1,882
1,791
ABM 16
0,564
0,331
2,444
5,259
5,645
0,831
5,645
1,217
-0,386
-0,386
18
-0,386
428,330
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
ABM 12
1,412
J1
1,725
IS
FS
1,319
COLL LINE
RL
AL
FL
430,131
428,719
428,719
428,719
430,537
428,812 -1
428,811
J2
1,527
429,010 -2
429,008
J3
1,318
429,219 -2
429,217
428,655 -2
428,653
428,986 -3
428,983
2,444
428,333 -3
428,330
428,330
5,645
428,719
AL
FL
J4
2,122
J5
1,882
430,777
1,791
ABM 16
5,259
5,645
-0,386
-0,386
Question 4
POINT
BS
TSM12
1,435
IS
2,296
80
3,527
0,274
FALL
RL
4,999
120
3,518
140
3,729
160
4,023
TSM39
RISE
505,273
60
100
FS
4,879
497,108
19
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
TSM12
1,435
IS
COLL LINE
RL
AL
FL
505,273
60
2,296
80
3,527
100
FS
0,274
4,999
120
3,518
140
3,729
160
4,023
TSM39
4,879
497,108
Question 5
POINT
BS
BM21
1,842
A
1,623
IS
MH1
1,322
E
1,224
BM22
RL
AL
FL
0,982
1,021
0,844
F
G
FALL
1,422
1,401
D
RISE
112,335
B
C
FS
1,002
0,996
1,823
1,228
2,044
113,002
20
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
BM21
1,842
A
1,623
IS
MH1
1,322
E
1,224
AL
FL
0,982
1,021
0,844
F
G
RL
1,422
1,401
D
COLL LINE
112,335
B
C
FS
1,002
0,996
1,823
1,228
BM22
2,044
113,002
Question 6
POINT
BS
BM22
1,329
P1
1,626
IS
P3
1,422
BM23
FALL
RL
AL
FL
1,482
1,343
P5
RISE
113,002
P2
P4
FS
2,041
1,463
1,824
2,308
112,751
21
SUTP101
SURVEYING: THEORY AND PRACTICAL
POINT
BS
BM22
1,329
P1
1,626
IS
RL
AL
FL
1,482
1,343
P3
1,422
2,041
P5
COLL LINE
113,002
P2
P4
FS
1,463
1,824
BM23
2,308
112,751
Question 7
POINT
BS
PEG1
1,997
INTA
IS
FS
RL
AL
FL
2,425
1,894
3,172
INTC
0,946
2,998
INTD
1,466
UND
-1,749
PEG6
FALL
243,597
INTB
INTE
RISE
1,942
3,467
2,949
237,798
22
.
POINT
BS
PEG1
1,997
INTA
IS
FS
RL
AL
FL
243,597
2,425
INTB
1,894
3,172
INTC
0,946
2,998
INTD
1,466
UND
-1,749
INTE
COLL LINE
1,942
3,467
PEG6
2,949
237,798
Question 8
POINT
BS
A
2,816
IS
2,287
C
1,849
D
1,236
2,314
FALL
RL
AL
FL
0,516
F
1,694
G
1,687
H
2,387
J
RISE
228,160
B
E
FS
3,148
229,632
23
POINT
BS
A
2,816
IS
COLL LINE
RL
AL
FL
228,160
B
2,287
C
1,849
D
1,236
E
FS
2,314
0,516
F
1,694
G
1,687
H
2,387
J
3,148
229,632
Question 9
POINT
BS
PI 17
2,746
IS
2,308
140
3,667
0,314
FALL
RL
AL
FL
2,999
180
3,618
200
2,994
220
4,023
PI 18
RISE
306,498
120
160
FS
3,576
302,987
24
POINT
BS
PI 17
2,746
IS
COLL LINE
RL
AL
FL
306,498
120
2,308
140
3,667
160
FS
0,314
2,999
180
3,618
200
2,994
220
4,023
PI 18
3,576
302,987
Question 10
POINT
BS
BM1
1,621
IS
0,200
T2
2,214
T3
1,700
1,552
T5
T6
FALL
RL
AL
FL
1,781
1,500
1,513
1,492
T7
1,513
T8
0,794
BM2
RISE
200,000
T1
T4
FS
1,656
199,763
25
POINT
BS
BM1
1,621
IS
0,200
T2
2,214
T3
1,700
1,552
T5
COLL LINE
RL
AL
FL
200,000
T1
T4
FS
1,781
1,500
T6
1,513
1,492
T7
1,513
T8
0,794
BM2
1,656
199,763
Question 11
POINT
BS
ABM 12
1,126
J1
1,878
IS
CULV
-1,338
ABM 16
FALL
RL
AL
FL
1,522
1,466
J5
RISE
28,719
J2
J4
FS
2,410
1,991
1,549
2,422
28,192
26
POINT
BS
ABM 12
1,126
J1
1,878
IS
FS
RL
AL
FL
28,719
1,522
J2
1,466
CULV
-1,338
J4
COLL LINE
2,410
1,991
J5
1,549
ABM 16
2,422
28,192
PTO FOR SOLUTIONS
Solutions: Question2
POINT
BS
TSM55
1,771
P
1,315
IS
1,619
Q
1,880
CP1
1,424
R
1,419
S
T
0,899
0,912
GATE
7,149
RL
Cor
AL
FL
242,782
0
242,782
242,782
FALL
242,934
+2
242,936
242,369
+4
242,373
0,456
242,825
+4
242,829
0,525
243,350
+4
243,354
0,313
243,663
+6
243,669
243,565
+6
243,571
243,592
+8
243,600
0,152
1,204
0,885
1,732
RISE
0,565
1,106
U
V
FS
0,098
0,027
1,321
0,436
243,156
+8
243,164
2,072
0,340
242,816
+10
242,826
1,439
242,782
7,115
1,473
7,115
1,439
0,034
0,034
0,034
27
242,826
POINT
BS
IS
TSM55
1,771
P
1,315
FS
1,619
COLL LINE
RL
Cor
AL
FL
244,553
242,782
0
242,782
242,782
244,249
242,934
+2
242,936
Q
1,880
242,369
+4
242,373
CP1
1,424
242,825
+4
242,829
243,350
+4
243,354
243,663
+6
243,669
243,565
+6
243,571
243,592
+8
243,600
243,156
+8
243,164
2,072
242,816
+10
242,826
242,826
7,115
242,782
RL
AL
FL
505,273
505,273
505,273
R
1,419
S
0,899
244,769
1,106
T
0,912
U
1,204
244,477
0,885
V
1,732
1,321
GATE
7,149
244,888
7,115
0,034
0,034
Question 4
POINT
BS
TSM12
1,435
IS
FS
RISE
FALL
60
2,296
0,861
504,412+2
504,414
80
3,527
1,231
503,181+2
503,183
1,472
501,709+2
501,711
100
0,274
4,999
120
3,518
3,244
498,465+4
498,469
140
3,729
0,211
498,254+4
498,258
160
4,023
0,294
497,960+4
497,964
0,856
497,104+4
497,108
8,169
505,273
TSM39
4,879
1,709
9,878
0
9,878
8,169
-8,169
-8,169
28
-8,169
497,108
POINT
BS
TSM12
1,435
IS
FS
COLL LINE
RL
AL
FL
506,708
505,273
505,273
505,273
60
2,296
504,412+2
504,414
80
3,527
503,181+2
503,183
501,709+2
501,711
100
0,274
4,999
501,983
120
3,518
498,465+4
498,469
140
3,729
498,254+4
498,258
160
4,023
497,960+4
497,964
4,879
497,104+4
497,108
497,108
9,878
505,273
RL
AL
FL
112,335
112,335
112,335
0,420
112,755-2
112,753
TSM39
1,709
9,878
-8,169
-8,169
Question 5
POINT
BS
BM21
1,842
A
1,623
IS
FS
1,422
RISE
FALL
B
1,401
0,222
112,977-4
112,973
MH1
1,322
0,079
113,056-4
113,052
0,340
113,396-4
113,392
0,203
113,599-7
113,592
0,019
113,618-7
113,611
0,152
113,466-9
113,457
1,228
0,232
113,234-9
113,225
2,044
0,221
113,013-11
113,002
0,605
112,335
C
1,224
D
E
1,021
0,844
F
G
0,982
1,002
0,996
1,823
BM22
7,356
6,678
1,283
6,678
0,605
0,678
0,678
29
0,678
113,002
POINT
BS
BM21
1,842
A
1,623
IS
FS
1,422
COLL LINE
RL
AL
FL
114,177
112,335
112,335
112,335
114,378
112,755-2
112,753
B
1,401
112,977-4
112,973
MH1
1,322
113,056-4
113,052
113,396-4
113,392
113,599-7
113,592
113,618-7
113,611
113,466-9
113,457
113,234-9
113,225
2,044
113,013-11
113,002
113,002
6,678
112,335
RL
AL
FL
113,002
113,002
113,002
112,849+2
112,851
113,132+4
113,136
0,079
113,053+4
113,057
0,041
113,012+4
113,016
113,229+6
113,235
0,484
112,745+6
112,751
0,757
113,002
C
1,224
D
E
114,620
1,021
0,844
F
G
0,982
1,002
114,462
0,996
1,823
1,228
BM22
7,356
115,057
6,678
0,678
0,678
Question 6
POINT
BS
BM22
1,329
P1
1,626
IS
1,343
P3
1,422
2,041
P5
0,153
1,463
0,217
2,308
4,996
FALL
0,283
1,824
BM23
RISE
1,482
P2
P4
FS
5,253
0,500
5,253
0,757
-0,257
-0,257
30
-0,257
112,751
POINT
BS
BM22
1,329
P1
1,626
IS
FS
COLL LINE
RL
AL
FL
114,331
113,002
113,002
113,002
114,475
112,849+2
112,851
1,482
P2
1,343
113,132+4
113,136
P3
1,422
113,053+4
113,057
113,012+4
113,016
113,229+6
113,235
2,308
112,745+6
112,751
112,751
5,253
113,002
RL
AL
FL
243,597
243,597
243,597
0,428
243,169+2
243,171
P4
2,041
P5
1,463
115,053
1,824
BM23
4,996
5,253
-0,257
-0,257
Question 7
POINT
BS
PEG1
1,997
INTA
IS
FS
RISE
2,425
FALL
INTB
1,894
3,172
0,747
242,422+2
242,424
INTC
0,946
2,998
1,104
241,318+4
241,322
0,520
240,798+6
240,804
244,013+6
244,019
INTD
1,466
UND
-1,749
INTE
1,942
PEG6
6,779
3,215
3,467
5,216
238,797+6
238,803
2,949
1,007
237,790+8
237,798
9,022
243,597
12,586
3,215
12,586
9,022
-5,807
-5,807
31
-5,807
237,798
.
POINT
BS
PEG1
1,997
INTA
IS
FS
COLL LINE
RL
AL
FL
245,594
243,597
243,597
243,597
243,169+2
243,171
2,425
INTB
1,894
3,172
244,316
242,422+2
242,424
INTC
0,946
2,998
242,264
241,318+4
241,322
INTD
1,466
240,798+6
240,804
UND
-1,749
244,013+6
244,019
238,797+6
238,803
2,949
237,790+8
237,798
237,798
12,586
243,597
RL
AL
FL
228,160
228,160
228,160
INTE
1,942
3,467
PEG6
6,779
240,739
12,586
-5,807
-5,807
Question 8
POINT
BS
A
2,816
IS
FS
RISE
FALL
B
2,287
0,529
228,689+3
228,692
C
1,849
0,438
229,127+3
229,130
D
1,236
0,613
229,740+3
229,743
0,720
230,460+3
230,463
E
2,314
0,516
F
1,694
0,620
231,080+6
231,086
G
1,687
0,007
231,087+6
231,093
H
2,387
0,700
230,387+6
230,393
0,761
229,626+6
229,632
1,461
228,160
J
3,148
5,130
3,664
2,927
3,664
1,461
1,466
1,466
32
1,466
229,632
POINT
BS
A
2,816
IS
FS
COLL LINE
RL
AL
FL
230,976
228,160
228,160
228,160
B
2,287
228,689+3
228,692
C
1,849
229,127+3
229,130
D
1,236
229,740+3
229,743
230,460+3
230,463
E
2,314
0,516
232,774
F
1,694
231,080+6
231,086
G
1,687
231,087+6
231,093
H
2,387
230,387+6
230,393
3,148
229,626+6
229,632
229,632
3,664
228,160
RL
AL
FL
306,498
306,498
306,498
306,936+2
306,938
305,577+2
305,579
306,245+2
306,247
302,941+4
302,945
303,565+4
303,569
302,536+4
302,540
302,983+4
302,987
J
5,130
3,664
1,466
1,466
Question 9
POINT
BS
PI 17
2,746
IS
120
2,308
140
3,667
160
0,314
3,618
200
2,994
220
4,023
3,060
RISE
FALL
0,438
1,359
2,999
180
PI 18
FS
0,668
3,304
0,624
1,029
3,576
0,447
6,575
2,177
6,575
5,692
-3,515
-3,515
33
5,692
306,498
-3,515
302,987
POINT
BS
PI 17
2,746
IS
FS
COLL LINE
RL
AL
FL
309,244
306,498
306,498
306,498
120
2,308
306,936+2
306,938
140
3,667
305,577+2
305,579
306,245+2
306,247
160
0,314
2,999
306,559
180
3,618
302,941+4
302,945
200
2,994
303,565+4
303,569
220
4,023
302,536+4
302,540
3,576
302,983+4
302,987
302,987
6,575
306,498
RL
AL
FL
200,000
200,000
200,000
201,421+2
201,423
199,407+2
199,409
199,921+2
199,923
199,840+2
199,842
0,052
199,892+4
199,896
0,008
199,900+4
199,904
199,900+6
199,906
200,619+6
200,625
0,862
199,757+6
199,763
2,957
200,000
PI 18
3,060
6,575
-3,515
-3,515
Question 10
POINT
BS
BM1
1,621
IS
T1
0,200
T2
2,214
T3
1,700
T4
1,552
T5
T6
FS
RISE
1,421
2,014
0,514
1,781
1,500
1,513
1,492
0,081
T7
1,513
0
T8
0,794
0,719
BM2
1,656
4,686
FALL
4,929
2,714
4,929
2,957
-0,243
-0,243
34
0
-0,243
199,763
POINT
BS
BM1
1,621
IS
FS
COLL LINE
RL
AL
FL
201,621
200,000
200,000
200,000
T1
0,200
201,421+2
201,423
T2
2,214
199,407+2
199,409
T3
1,700
199,921+2
199,923
199,840+2
199,842
199,892+4
199,896
199,900+4
199,904
T4
1,552
T5
T6
1,781
201,392
1,500
1,513
1,492
201,413
T7
1,513
199,900+6
199,906
T8
0,794
200,619+6
200,625
1,656
199,757+6
199,763
4,929
200,000
BM2
4,686
4,929
-0,243
-0,243
35
199,763
SUTP101
SURVEYING: THEORY AND PRACTICAL
Question 11
POINT
BS
ABM 12
1,126
J1
1,878
IS
FS
RISE
1,522
FALL
0,396
RL
AL
FL
28,719
28,719
28,719
28,323-2
28,321
J2
1,466
0,412
28,735-4
28,731
CULV
-1,338
2,804
31,539-4
31,535
28,210-4
28,206
29,071-6
29,065
0,873
28,198-6
28,192
28,192
4,598
28,719
J4
2,410
J5
1,991
1,549
ABM 16
3,329
0,861
2,422
5,414
5,935
4,077
5,935
4,598
-0,521
-0,521
POINT
BS
ABM 12
1,126
J1
1,878
IS
FS
1,522
-0,521
COLL LINE
RL
AL
FL
29,845
28,719
28,719
28,719
30,201
28,323-2
28,321
J2
1,466
28,735-4
28,731
CULV
-1,338
31,539-4
31,535
28,210-4
28,206
29,071-6
29,065
2,422
28,198-6
28,192
5,935
28,719
J4
2,410
J5
1,549
ABM 16
5,414
5,935
-0,521
Page i of 285
1,991
30,620
-0,521
28,192
SURV111/108
SURVEYING1 CIVIL
Chapter 4
DISTANCE AND ANGLE MEASUREMENT
Note: this chapter will cover reducing measured distances to reduced
distances. When setting out, the surveyor has reduced distances available and
will have to convert those to the equivalent measured distances for setting out
4.1
Chain
A chain consists of 100 main wire or metal links, with three small ring links between
each for flexibility. Brass handles are attached to the ends and the nominal length of
a chain is measured to the outside of the handles.
Brass tallies at every tenth link show how many tens of links between the tally and
the nearest handle. A common source of error in chaining is in misreading the units.
The two main types of chains are the 66ft
Engineer's chain.
long Gunter's chain and the 100 ft
The chain is robust and, if broken, easily repaired. However, being heavy, clumsy to
use and not very accurate, it has been superseded in survey work by the Tape.
4.2
Metallic and Linen Tapes
These types may or may not be plastic coated. The linen variety is made of woven
linen with painted graduations. This type stretches very easily with about 2% or even
more stretch. The metallic variety has copper strands woven into the linen, resulting
in less stretch. Metallic tapes should not be used near exposed electrical wiring or
conductors. These tapes are available in lengths of 10m, 15m, 20m, 25m and 30m,
and graduated in metres and centimetres.
4.3
Steel Tapes and Bands
These are lighter, more accurate but much more susceptible to damage. The type
most suitable for engineering work is the 6mm wide band in lengths of 30m, 50m or
100m. The lengths generally exclude the handles which are usually detachable. The
band is kept wound on a steel cross frame fitted with a winding handle.
Some tapes are graduated in metres and hundredths of a metre on one side.
The operation of measuring is still referred to as chaining a distance. Chaining is one
of the basic and most important operations in survey work.
Some Advantages of the Steel Tape
1.
2.
3.
It is light
It maintains its length accurately
It has accurate graduations
SURV111/108
SURVEYING1 CIVIL
4.
It may be stored on a reel and long lengths are handled with ease.
Some Disadvantages
1.
2.
3.
4.
4.4
Being thin and light, if carelessly used, will break.
Not easily repaired and any repair will affect the accuracy.
To preserve legibility must be kept free from rust.
Too great a tension will give the tape a "set" which permanently affects
the accuracy.
Care of Steel Tapes
Steel tapes rust and, before being put away for any length of time and always if wet,
must be dried and wiped over with an oily rag. Metal polishes or abrasives must
not be used since they affect the legibility and even the weight of the tape.
Steel tapes stretch easily but return to their original length unless stretched too far. A
safe tension (or pull) is that which can be applied using one hand, without
throwing the weight of the body into the pull.
Being brittle they break easily. They must always be pulled out in line with the
opening; that is more or less tangential to the reel. Before applying tension to
straighten the tape, make sure there are no kinks in it.
Do not allow cars to ride over the tape and avoid stepping on it while on the
ground. Never leave the tape lying on the ground longer than necessary as this
increases the danger to damage.
Stainless steel tapes do not rust and are not as brittle as the carbon steel tapes.
Steel tapes are also available with a vinyl coating having sharp black graduations on
white or yellow background.
4.5
Corrections to Measured Lengths
The following corrections must be applied, when measuring with a steel tape, to give
the correct horizontal distance between points:
1.
2.
3.
4.
5.
Standard
Temperature
Tension
Slope
Sag
Note: The corrections above are applied to all distances measured by tape. If such
distances are to be used in coordinate calculation on the South African
Coordinate system, the corrections for Mean Sea Level and Scale
Enlargement have to be applied in addition to those above.
In this course we will not be applying these so-called “projection corrections”.
SURV111/108
SURVEYING1 CIVIL
4.5.1 Standard
The tapes are supplied by the maker as being of a standard length, at a
particular temperature and tension. A brass tag attached to the handle or the
frame usually states the temperature and tension under which the tape is of
correct length. As tapes are sometimes stretched they should be periodically
checked against a standard length, and a new standard value obtained for
them.
Note: Provided the tension under which the tape was standardized is
used when a temperature correction is applied, this automatically
also applies the standard correction.
4.5.2 Temperature
A tape extends with a rise in temperature and contracts with a fall in
temperature. The correction is dependent on the coefficient of expansion of
the steel tape. This may vary but a mean expansion of 0,000012 per 100 can
be used in the formula:
Ct
=
L x e(tm - ts)
Where: Ct
L
e
tm
ts
=
=
=
=
=
the correction for temperature
measured length to be corrected
coefficient of expansion of steel
temperature during measurement
temperature at which standardized
NOTE: THE SIGN OF THE CORRECTION is dependent on the SIGN OF THE
EXPRESSION (tm - ts).
4.5.3 Tension
For lightness, steel tapes are made of as narrow a cross-section as possible.
This means they are sensitive to variation in tension, and for accurate
measurements the standard tension should be maintained by means of a
spring balance.
When using a spring balance an essential item of equipment is a "Little John
Roller Grip", which is a small but ingenious device enabling the spring balance
to be attached to the tape at any point along its length.
A fairly strong pull is necessary to straighten kinks, but too heavy a pull might
cause a permanent "set". A tension of about 70 newtons is sufficient to ensure
accuracy and will not damage the tape.
SURV111/108
NOTE:
SURVEYING1 CIVIL
The correction for tension is seldom applied, this being taken
care of by use of the correct tension.
4.5.4 Slope
In most survey work, and especially for the drawing of plans, the horizontal
distance is required, so that distances measured on a slope have to be
reduced to the horizontal.
C
=
AB(1 - cos α)
For reducing measured distances to reduced distances, this correction
is always negative since the horizontal distance is always less than the
slope distance.
4.5.5 Sag
There are two recognised methods of taping distances accurately:
i.
Taping along the ground. This is only possible when the ground is clear
of obstructions and slopes evenly.
To measure the slope of the ground, measure the height of the
instrument and mark that height on a staff or rod. Therefore, a
parallel line to the slope of the ground is obtained.
ii.
When the above conditions are not met "taping in catenary" is used. In
this method the tape is suspended above the ground. The tape then
naturally sags and the curve formed is similar to that of a suspended
chain, hence the name (catenary) from the Latin "catina" - a chain.
If the tape was standardized on a level surface and it is used in
catenary, it is then necessary to apply a correction to obtain the chord
length, as opposed to the arc length. The correction is always negative
when measuring.
SUTP101
SURVEYING: THEORY AND PRACTICAL
Cc
=
w²L³
24T²
Where: Cc = the sag correction also known as catenary correction
w = the weight of the tape in Kg per metre run
L = length of the tape under consideration
T = pull (or tension) applied in Kg
4.6
Errors and Mistakes in Taping
It is not possible to determine the exact or true length of a line or, for that matter, the
magnitude of an angle. The values obtained by physical measurement are only
relative to the unit of measure and invariably subject to error, which can be defined
as the difference between the true and the measured value of a quantity.
The different types of error can be classified into:
Mistakes, Constant Errors, Systematic Errors, Accidental Errors
TYPICAL MISTAKES in taping a line are:
1.
2.
3.
Forgetting to book a distance
Misreading the tape
Not booking the distance measured
CONSTANT AND SYSTEMATIC errors arise from sources which are known, or
should be known, and their effects can be eliminated. For example, a tape that has
been broken and subsequently repaired so that it is not of correct length, will give a
constant error.
SYSTEMATIC errors are a result of temperature differences, height above sea level,
etc. Corrections can be calculated and applied to them.
4.7
Calculating the corrections
Example 1: Complete the table below. The tape used was standard at 25°C. Use a
coefficient of expansion of 0,000012/°C. The weight of the tape was
0,015kg/m.
POINTS
A-B
MEASURED
DISTANCE
264,76m
MEASURED
TEMP
33°C
CS=AB(1 - cos α)
FORMULAE:
SLOPE CORR
-0,471
TEMP CORR
+0,025
Page i of 285
SLOPE
ANGLE
+3°25'
NUMBER OF BAYS
On ground
Ct=L x e(tm - ts)
SUTP101
CIVIL ENGINEERING I
TOTAL CORR
-0,445
REDUCED DIST
264,315
MEASURED
DISTANCE
SLOPE
ANGLE
MEASURED
TEMP
NUMBER OF BAYS
99,56m
0
25°C
Tape supported at
the 50m graduation
POINTS
F–G
Cc= w²L³
24T²
FORMULA:
SAG CORR 1 =
-0,024
SAG CORR 2 =
-0,023
TOTAL CORR =
-0,047
REDUCED DIST =
99,513
Example 2: Complete the tables below. The tape used was standard at 20C. Use a
coefficient of expansion of 0,000012/C. The weight of the tape was 0,015kg/m.
Standard tension of 7kg was used throughout
POINTS
A-B
MEASURED
DISTANCE
284,135m
-347'
MEASURED
TEMP
NUMBER OF BAYS
11C
On ground
Ct=L x e(tm - ts) C=AB(1 - cos α)
FORMULAE:
SLOPE CORR =
- 0,619m
TEMP CORR =
- 0,031m
TOTAL CORR =
- 0,650m
REDUCED DIST =
283,485m
F-G
SLOPE
ANGLE
96,653m
0
FORMULA:
20C
Cc= w²L³
24T²
SAG CORR 1 =
FOR 40m : -0,012m
SAG CORR 2 =
FOR 56,653m : -0,035m
TOTAL CORR =
-0,047m
REDUCED DIST =
96,606m
ii
Tape supported at
the 40m graduation
SUTP101
CIVIL ENGINEERING I
Example 3: Complete the table below. The tape used was standard at 25°C. Use a
coefficient of expansion of 0,000012/°C. The weight of the tape was 0,015kg/m.
Standard tension of 7kg was used throughout.
POINTS
MEASURED
DISTANCE
A-B
SLOPE
ANGLE
195,424
+3°45'
MEASURED
TEMP
NUMBER OF BAYS
33°C
On ground
C t = L x e(tm - ts) C = AB(1 - cos α)
FORMULAE:
SLOPE CORR =
-0,418
TEMP CORR =
+0,019
TOTAL CORR =
-0,399
REDUCED DIST =
195,025
F-G
79,23m
0
FORMULA:
25°C
Tape supported at
the 50m graduation
Cc= w²L³
24T²
SAG CORR 1 =
-0,024
SAG CORR 2 =
-0,005
TOTAL CORR =
-0,029
REDUCED DIST =
79,201
Exercise
Complete the tables below. The tape used was standard at 20C. Use a coefficient
of expansion of 0,000012/C. The weight of the tape was 0,015kg/m. Standard
tension of 7kg was used throughout.
POINTS
A-B
MEASURED
DISTANCE
284,135m
SLOPE
ANGLE
-347'
FORMULAE:
SLOPE CORR =
TEMP CORR =
TOTAL CORR =
iii
MEASURED
TEMP
NUMBER OF BAYS
11C
On ground
SUTP101
POINTS
CIVIL ENGINEERING I
MEASURED
DISTANCE
SLOPE
ANGLE
MEASURED
TEMP
NUMBER OF BAYS
20C
Tape supported at
the 40m graduation
REDUCED DIST =
F-G
96,653m
0
FORMULA:
SAG CORR 1 =
SAG CORR 2 =
TOTAL CORR =
REDUCED DIST =
4.8
Optical Distance Measurement
A typical surveyor's theodolite has a pair of stadia marks.
The stadia marks are set a specific distance apart. The distance is chosen so that
there is a fixed, integer ratio between the distance observed between the marks and
the distance from the telescope to the measuring device observed. This is known as
the stadia constant or stadia interval factor. For example, a typical vertical stadia
mark pair is set so that the ratio is 100. If one observes a levelling staff with the
telescope and sees that the readings spans 0.5m between the marks (the stadia
interval), then the distance from the instrument to the staff is:
0,5m x 100 = 50 m
when the line of sight is completely horizontal
In the image below, the upper stadia mark is at 1,5 m and the lower at 1,345 m. The
difference is 0,155 m. Thus the distance from the instrument to the levelling staff,
when the line of sight is completely horizontal, is:
0,155 x 100 = 15.5 m
iv
SUTP101
CIVIL ENGINEERING I
When the line of sight is not horizontal, the distance is:
100S Cos²α where α is the measured vertical angle and S is the stadia interval
OR 100S Sin²Z where Z is the zenith angle (vertical reading) on the theodolite
If for the readings above, Z = 87°, which means α = +3°, then the horizontal distance
would be:
100x0,155 Cos² 3° = 15,458m OR 100x0,155Sin² 87° = 15,458m
4.9
Electronic Distance Measurement (EDM)
An electronic instrument placed at one end of a line transmits electromagnetic waves
to a retro-reflector positioned at the other end of the line. The reflector returns the
waves to the transmitter. The instrument computes the length of the line and displays
it as a digital read-out.
As the speed of the waves is known the distance can be determined, by measuring
the time interval between emission and return.
In practice, the instrument transmits signals at several different frequencies,
measures the phase shift between the emitted and reflected signal at each
frequency, then from a comparison of the phase shift, together with knowledge of the
different wave lengths used, deduces the distance.
As the velocity of the waves depends on the prevailing atmospheric conditions
(temperature, humidity, etc.,) a correction for deviation from the calibrated conditions
must be applied. Slope, sea level and scale corrections must also be considered.
A variety of instruments using radio waves, visible light, infrared radiation or laser
beams, are made use of. The maximum range varies from 100+ Km to several
hundred metres.
v
SUTP101
4.10
CIVIL ENGINEERING I
EDM Corrections
Note: The corrections below are applied to all distances measured by EDM. If such
distances are to be used in coordinate calculation on the South African
Coordinate system, the corrections for Mean Sea Level and Scale
Enlargement have to be applied in addition to those above.
In this course we will not be applying these so-called “projection
corrections”. Students have to be aware that such corrections are essential
when doing work on the SA Coordinate system, and will be covered in the S1
– S4 attendances.
Slope Correction
When the slope distance has been obtained from an EDM measurement, a slope
correction must be applied to it in order to obtain the equivalent horizontal distance.
If the EDM and theodolite are coaxial as in most total stations and the telescope is
tilted and pointed at the centre of the prism the correct slope distance is obtained no
matter how the prism is tilted and the horizontal distance can be calculated. For an
integrated total station, this calculation is carried out by the instrument's
microprocessor and the result displayed automatically.
Meteorological Corrections
Using EDM equipment, the measurement of distance is obtained by measuring the
time of propagation of electromagnetic waves through the atmosphere. Whilst the
velocity of these waves in a vacuum is known, its value will be reduced according to
the atmospheric conditions through which the waves travel at the time of
measurement.
The value of the correction is affected by the temperature, pressure and water
vapour content of the atmosphere as well as by the wavelength of the transmitted
electromagnetic waves. It follows from this that measurements of these atmospheric
conditions are required at the time of measurement.
EDM equipment is standardized under certain conditions of temperature and
pressure. For instance, some makes of equipment are standardized at 20°C and
1013.25 mbar of pressure, whilst others are standardized at 12°C and 1013.25 mbar.
Under these respective atmospheric conditions, the measured distances would not
require a velocity correction. It follows that even on low-order surveys the
measurement of temperature and pressure is important.
Instrument & Prism Constant, also known as the Zero Error (Z)
This occurs if there are differences in the mechanical, electrical and optical centres
of the EDM instrument and reflectors, and includes the prism constant. It is therefore
due to a difference between the mechanically defined centres of instruments (and
reflector-when used) and their electrical (optical) centres. The error when present,
and not allowed for, produces an effect akin to a mis-centering of the instrument by
the operator and is independent of range.
vi
SUTP101
CIVIL ENGINEERING I
This is therefore composed of two parts that cannot easily be distinguished;
(a)
The distance between the plumbing centre of the EDM and its “electrical
centre”; i.e. the point from which distances are effectively measured. This
value is kept small by the EDM design.
(b)
The distance from the prism plumbing centre, to the point from which the
signal is reflected back to the EDM. This value depends on the make of prism
and is generally between -0,03m and -0,08m
This error is of constant magnitude, and care must be taken to eliminate it. The value
of a zero error obtained from a calibration procedure usually applies to an instrument
and reflector and if the reflector is changed, the zero constant changes.
It is important to note that the zero (Z) value is usually obtained by performing a base
line measurement in the field. Some handbooks prescribe a procedure that will result
in the ERROR being calculated, whist others will result in the CORRECTION being
obtained.
EXAMPLES
1.
Reduce the given distances which were measured with an EDM.
Corrections for meteorological effects, slope and the instrument
constant must be applied.
List of Formulae:
•
Meteorological correction: +20mm /1000m (ALSO KNOWN AS 20
PARTS PER MILLION or PPM)
•
Slope: C = S(1 - cosα )
•
Correction for Instrument Constant: -0,142m
vii
SUTP101
CIVIL ENGINEERING I
SOLUTION
FIELD DATA
CORRECTIONS
STA
MEAS.
DIST.
VERT.
ANGLE
MET
SLOPE
INST
CONST
RED
DIST
A-B
264,491
2°11’
+0,005
-0,192
-0,142
264,162
B-C
201,894
3°01’
+0,004
-0,280
-0,142
201,476
C-D
5,072
11°17’
0
-0,098
-0,142
4,832
D-E
302,838
1°56’
+0,006
-0,172
-0,142
302,530
2.
Reduce the measured EDM distances
Formulae:
▪
▪
▪
Meteorological correction: 38mm per 1000m or 38ppm
Slope: C = S(1 - cos )
Error due to Instrument Constant: + 0,034m
FIELD DATA
CORRECTIONS
Reduced
Distance
Measured Distance
Vert
Angle
Slope
Corr
Met
Corr
Inst
const
2 122,664m
2°15'
-1,637
+0,081
-0,034
2121,074m
3 075,428m
2°30'
-2,927
+0,117
-0,034
3072,584m
1 896,431m
1°45'
-0,885
+0,072
-0,034
1895,584m
EXERCISES
1.
Reduce the measured EDM distances.
Measured slope distances and mean vertical angles appear below. The EDM
instrument has an Error due to the Instrument Constant of +0.008 and an
atmospheric correction of +20mm/1000m (20ppm). Reduce the distances to the
horizontal.
viii
SUTP101
CIVIL ENGINEERING I
Measured
Distance
Vert
Angle
Slope
Corr
Met
Corr
Inst
const
Reduced
Distance
dd.mm.ss
1169,679m
3.54.30
741,268m
5.49.25
1003,138m
6.31.15
994,563m
7.10.40
2.
An EDM device was used to measure twelve distances. Calculate the reduced
distances of the measured distances below, applying all the necessary
corrections.
o Meteorological correction: +10mm per 1000m (10ppm)
o Slope: C = S(1 - cos )
o Correction for Instrument Constant: - 0,004m
Measured
Distance
4.11
Vert
Angle
dd.mm.ss
852,383m
2.13.46
641,574m
7.19.33
1 009,792m
8.41.07
878,221m
4.42.21
724.863m
2.22.52
868,419m
1.00.00
75,322m
5.16.33
5,000m
0.00.00
9947,462m
5.21.26
1657,531m
11.11.11
5454,333m
00.05.05
1627,477
01,42,26
Slope
Corr
Met
Corr
Inst const
Reduced
Distance
Mechanical construction and angle measurement
The main features of the electronic angle measuring theodolite are discussed below,
and shown in the Figure on page 62.
4.11.1 The three axes
ix
SUTP101
CIVIL ENGINEERING I
The standing or vertical axis is what the instrument revolves around. When setting
up, the surveyor sets this axis vertical.
The trunnion or horizontal axis is what the telescope tilts around. The
manufacturer makes this at right angles to the standing axis. It is also called the
tilting axis and horizontal axis.
The line of sight or line of collimation is the pointing line. It passes through the
intersection of standing and trunnion axes and is meant to be at right angles to the
trunnion axis.
4.11.2 The two circles
The observed direction and zenith angle are sensed on the horizontal and vertical
axes respectively.
The horizontal circle can be rotated or oriented with respect to the survey grid by
keying in the true oriented direction when the line of sight is towards a target with
known direction on the grid. This could be the join direction.
The vertical circle is set so that when the line of sight is vertically upwards towards
the zenith, the reading is zero.
4.11.3 The index compensator is a gravity-direction sensor that accurately adjusts
the vertical circle orientation when the standing axis is not set up accurately vertical.
4.11.4 The plate bubble is a sensitive device for levelling up the standing axis. In
older instruments it is a tubular spirit vial. If it is simply a circular bubble then the
instrument has a dual-axis compensator. This digital sensor measures the nonverticality of the standing axis in two component directions.
•
Longitudinal dislevelment, away or towards the observer in the plane of the
line of sight. This is the index compensator mentioned above.
•
Transverse dislevelment left or right in the plane containing the trunnion axis.
This measured dislevelment is used to correct the horizontal circle reading.
x
SUTP101
CIVIL ENGINEERING I
Components of a Theodolite
4.11.5 The tribrach is the base that sits on the tripod. It has three levelling screws.
Usually the instrument can be lifted off the tribrach and replaced with a target or
other instrument.
4.11.6 The plummet is used for centring over the set-up point. It may have the form
of a visible laser plummet that shines down the standing axis. Another form is the
optical plummet, which is a little telescope with cross-hairs that sights down this
axis. The optical plummet may be attached to the tribrach. That allows the assistant
to centre and level a tribrach before the surveyor arrives with the total station.
4.12
Basic setting-up and observing skills
The basic skills are acquired by practicing: setting up the tripod over a ground point,
final levelling up and observing.
See your PRACTICAL MANUAL for details
Chapter 5
xi
SUTP101
CIVIL ENGINEERING I
TRAVERSING
A traverse consists of measurements in the field being carried out between successive points,
starting and ending at points with known coordinates. Once the measurements in the field
have been completed, a traverse calculation and adjustment are done to obtain coordinates
for the unknown points.
5.1
Traverse types
5.1.1 Closed Loop Traverse
Starts & closes at the same known point
5.1.2 Closed Route Traverse
Starts at known point and closes at different known point
5.2
Traverse field work
•
•
•
•
•
Reconnaissance: always plan the route of a traverse, starting from known point
and finishing on known point (unless not possible)
Follow correct observation procedure
Book all the information including what may seem insignificant, e.g. date.
Use well trained assistants familiar with the routine and are able to set up targets
Avoid short traverse legs
xii
SUTP101
CIVIL ENGINEERING I
5.2.1 Orientation
Once the theodolite has been set up over the starting point of the traverse – a
point with known coordinates – calculate joins to at least two other visible
known points for orientation. Select the longest ray, which in addition is well
defined, as your reference object (RO), and bisect the RO. Then follow the
prescribed procedure for the make/ model of theodolite you are using to set
the join direction into the instrument.
Check your orientation by bisecting the other known point which is visible from
your instrument position. Compare this direction with the join direction, which
should not differ by more than a few seconds.
5.2.2 The observation procedure (manual booking)
All the points in a clockwise direction are observed. The results are recorded
in the field book starting with the RO and closing again on the same point. It is
usual practise to do the first round of observation with the instrument in the
“Face Left” position (also known as “Circle Left”).
On completion of the CL observations, transit the telescope by turning it in the
vertical plane and rotate the instrument through 180 degrees. Re-observe the
RO, complete the remainder of the CR observations and book the results
starting from the bottom and working to the top.
5.2.3 Reduction of the field book
Once the observations have been booked, they must be meaned as follows:
Station
CL
CR
Mean
Corr.
Adj. Direction
(not oriented)
B
70 14 15
250 14 25
70 14 20
0
70 14 20
T1
140 12 38
320 12 46
140 12 42
-02
140 12 40
C
310 18 34
130 18 44
310 18 39
-04
310 18 35
RO
70 14 22
250 14 30
70 14 26
-06
@A
B and C were used for orientation, and T1 is the first unknown traverse point.
xiii
SUTP101
CIVIL ENGINEERING I
EXERCISES.
1.
Note – instrument below has horizontal collimation (line of sight) error.
Four known points (A, B, D and E) were used for orientation. Trav1 is the
first unknown traverse point.
Station
CL
CR
Mean
Corr.
Adj. Dirn
A
80.59.28
261.02.14
81.00.51
0
81.00.51
B
154.41.21
334.44.05
154.42.43
-01
154.42.42
Trav1
203.44.29
23.47.23
203.45.56
-01
203.45.55
E
206.30.30
26.33.16
206.31.53
-02
206.31.51
D
294.03.00
114.05.46
294.04.23
-02
294.04.21
RO
80.59.37
261.02.12
81.00.54
-03
@CABLE HILL
2.
Note – instrument below has horizontal collimation (line of sight) error.
Four known points (MTW, ISP, EL and DH) were used for orientation. P1
is the first unknown traverse point.
@IFAFA
CL
MTW 171.04.43
ISP
244.46.36
P1
293.49.53
EL
296.35.48
DH
24.08.16
RO
171.04.43
5.3
CR
351.07.27
64.49.19
113.52.36
116.38.33
204.10.58
351.07.29
The traverse direction sheet
Note that the direction sheet makes use of joins to obtain a provisional orientation correction
at the starting and closing points of the traverse. For intermediate stations, the surveyor has to
rely on back orientation (to the previous traverse station), or where s/he could rely on
observations to external control points at the intermediate traverse stations.
In the example below no external control points were observed from the intermediate traverse
stations T1 and T2. At those points the surveyor therefore had to rely on back orientation (to
the previously occupied point) in the field. In this course only examples where back
orientation was used will be covered.
xiv
SUTP101
CIVIL ENGINEERING I
Step 1:
From the field book enter the station names and the reduced field
observations into columns 1 and 2.
Step 2:
At the known points (starting and closing points) enter the join
directions in column 7 and underline. At the starting point, find the
correction [7-2] to be applied to the observed directions. Enter these
corrections into column 6. Find the mean correction using all the rays.
Apply this correction to the remaining rays end enter into column 5.
This value gives the oriented forward (outgoing) direction to the first
intermediate (unknown) traverse station. This oriented outgoing ray
will now become the incoming ray at the first intermediate traverse
station.
Step 3:
Treat the observations at the traverse stations in sequence. Enter the
oriented incoming ray from the previous station into column 3. Find
the orientation correction [3-2] and enter into column 4. Apply this
correction to the value in column 3 to give an oriented outgoing ray to
the next traverse station and enter into column 5. This oriented
outgoing ray will now become the incoming ray at the next traverse
station. REPEAT THIS PROCEDURE UP UNTIL THE
PENULTIMATE STATION, I.E. THE LAST UNKOWN
TRAVERSE STATION BEFORE THE CLOSING POINT.
Step 4:
We now have an oriented outgoing ray from the penultimate station to
the known (closing) point on which the traverse closes. We also have
an oriented outgoing ray from the closing point to the penultimate
station, which we obtained as we did for the starting point in Step 2.
The procedure to be followed at the closing station is therefore
identical to the procedure we followed at the starting point.
The difference between the outgoing and incoming rays at the closing
station is the closing error and will be distributed proportionally at
each station.
PTO FOR STEP-BY-STEP EXAMPLE OF A TRAVERSE DIRECTION
SHEET WHERE ONLY BACK ORIENTATION WAS USED AT THE
UNKNOWN TRAVERSE STATIONS.
xv
SUTP101
CIVIL ENGINEERING I
STEP 1
1
2
@A
B
T1
C
70 14 20
140 12 40
310 18 35
@ T1
A
T2
320 12 50
165 17 53
@ T2
T1
D
345 17 37
150 00 04
@D
E
F
T2
06 27 44
142 18 28
330 00 20
3
4
xvi
5
6
7
SUTP101
CIVIL ENGINEERING I
STEP 2
1
@A
B
T1
C
2
3
4
70 14 20
140 12 40
310 18 35
5
6
7
+10
70 14 30
+08
310 18 43
49
+09
@ T1
A
T2
320 12 50
165 17 53
@ T2
T1
D
345 17 37
150 00 04
@D
E
F
T2
06 27 44
142 18 28
330 00 20
+12
+18
35
+15
xvii
06 27 56
142 18 46
SUTP101
CIVIL ENGINEERING I
STEP 3
1
@A
B
T1
C
2
3
4
70 14 20
140 12 40
310 18 35
5
6
7
+10
70 14 30
+08
310 18 43
49
+09
@ T1
A
T2
320 12 50
165 17 53
49
@ T2
T1
D
345 17 37
150 00 04
52
@D
E
F
T2
06 27 44
142 18 28
330 00 20
-01
52
+15
19
+12
+18
35
+15
xviii
06 27 56
142 18 46
SUTP101
CIVIL ENGINEERING I
STEP 4
1
@A
B
T1
C
2
3
4
70 14 20
140 12 40
310 18 35
5
6
7
49
+10
+04
+08
70 14 30
140 12 53
310 18 43
+09
@ T1
A
T2
320 12 50
165 17 53
49
@ T2
T1
D
345 17 37
150 00 04
52
@D
E
F
T2
06 27 44
142 18 28
330 00 20
-01
52
+08
165 18 00
19
+12
150 00 31
35
+12
+18
-04
06 27 56
142 18 46
330 00 31
+15
+15
Closure = (330).00.35 – (150).00.19 = +16
Correction = +16/4 = +04” per set-up
At the closing station the sign of the correction at the first
set-up can be reversed and entered at the closing station, in
this case -04”
xix
SUTP101
5.4
CIVIL ENGINEERING I
The Traverse Calculation
Use the adjusted directions from the direction sheet, and reduced distances to calculate the
coordinates of the intermediate traverse stations. This is accomplished by doing successive
Polars, writing down only the

Y and X values on the traverse calculation sheet.

The sum of
Y and
X should agree with the Y and X differences between the starting
and closing points, in this case between A and D. Various ways are used to distribute the
closing error but we will use the Bowditch Rule, which states:
Correction to Y difference =
Y closing error
x LENGTH OF LEG
Length of Traverse
Correction to X difference =
X closing error
x LENGTH OF LEG
Length of Traverse
Linear accuracy =
(Yclosing error )2 + (X closing error) 2
FROM THE REGULATIONS PROMULGATED IN TERMS OF SECTION 10 OF THE LAND SURVEY
ACT, 1997 (ACT No. 8 OF 1997)
Regulation 5
(a)
when the position of a point is determined by polars, traverse, triangulation,
trilateration, GPS or a combination of these methods, the displacement between any
observed ray, measured distance or GPS vector and the equivalent quantity derived
from the final co-ordinates of the point fixed shall not exceedfor Class A : A metres;
for Class B : 1,5A metres;
for Class C : 3A metres;
where A is equal to-
xx
SUTP101
CIVIL ENGINEERING I
and S is the distance between the known and the unknown point……………….. provided
further that in the case of a traverse the comparison is made to the misclosure of the traverse,
where S is the total length of the traverse in metres;
Traverse Calculation
Dir/ Dist
Join
Y
X
(students)
140 12 53
+130,740
-157,002
204,31
+0,040
+0,053
165 18 00
+44,299
-168,856
174,57
+0,034
+0,045
150 00 31
+91,921
-159,267
183,89
+0,036
+0,047
S=562,77
+266,960
-485,125
+0,110
+0,145
Station
Y
X
A
-7345,27
+3927,58
T1
-7214,49
+3770,63
T2
-7170,16
+3601,82
D
-7078,20
+3442,60
+267,07
-484,98
Linear closing error = 0,182m. Class A = 0,059m. Closure is below Class C.
For a further example of the traverse direction sheet using back orientation, PTO
( 0,145 / 562,77) *174,57
xxi
SUTP101
1
CIVIL ENGINEERING I
2
3
4
5
6
7
@Woodmead
Albert Δ
56.25.20
+22
56.25.42
Umkomaas Δ
29.19.10
+18
29.19.28
Trav1
126.12.20
+09
126.12.49
40
40
20
@Trav1
Woodmead
306.12.30
Trav2
15.28.42
40
10
52
+19
15.29.11
@Trav2
52
02
Trav1
195.28.50
Umhlatuzana
14.56.20
22
+28
14.56.50
Trav2
194.56.27
59
-09
194.56.50
HCT Δ
126.29.29
+29
126.29.58
Phoenix Δ
243.55.49
+33
243.56.22
Worlds View Δ
319.59.59
+34
320.00.33
@ Umhlatuzana
+96
+32
Closure = (194).56.59 – (194).56.22 = 37
Correction = 37/4 = 9,25” per set-up
9,25x1 = 09’’ at Woodmead – enter into column 6
9,25x2 = 19” at Trav1
etc
At the closing station the sign of the correction at the first
set-up can be reversed and entered at the closing station, in
this case -09”
xxii
SUTP101
CIVIL ENGINEERING I
EXERCISES
1.
Obtain oriented directions for the calculation of final
co ordinates for T1 and T2 from the information given.
Do not calculate the traverse.
1
2
3
4
5
6
7
@ Begin
Berg
196. 03. 07
+07
196.03.14
Blou
83. 36. 56
+08
83.37.04
T1
159. 21. 30
+17
159.21.55
38
+15
+08
@ T1
T2
110.20.26
Begin
339.21.18
38
+12
30
+33
339.22.03
34
+50
139.02.24
@ T2
T1
290.19.54
End
139.01.58
30
-24
@ End
Berg
201. 55. 23
+20
201.55.43
Blou
79. 20. 23
+20
79.20.43
Begin
317. 08. 24
+20
317.08.44
T2
319.02.20
-17
319.02.23
40
+60
+20
319.02.40 - 319.01.34 = 0.01.06/4 = 16,5
xxiii
SUTP101
2.
CIVIL ENGINEERING I
Calculate co-ordinates for D, E and F by a traverse calculation with the following
reduced data:
Oriented Directions Reduced Distances
AD
DE
EF
FB
169.10.30
184.50.20
201.00.10
229.19.40
Co-ordinates A
3.
316,29m
422,37m
376,26m
309,84m
-13 789,45
+14 258,36
B
-14 135,67
+12 973,64
Calculate co-ordinates for T1, T2 and T3 by a traverse calculation with the following
reduced data:
Oriented Directions Reduced Distances
S – T1
T1 – T2
T2 – T3
T3 – T
291.30.10
303.30.28
02.30.36
08.35.04
Co-ordinates S
248,29m
199,49m
200,82m
227,70m
+13 142,91 +6295,84
T
+12788,34
+6922,77
SOLUTIONS FOR NO’S 2 AND 3 (COMPUTER PRINTOUT)
2.
Traverse Adjustment using the Bowditch Proportional [DY, DX] Method
Adjusted Data Traverse between Points A and B
Observed
Final
Data
Data
Name
Y
X
——————————————————————————————————————————————————————————————————————
A
-13789,450
+14258,360
169:10:30,0
169:10:41,3
316,290
316,286
D
-13730,065
+13947,699
======================================
184:50:20,0
184:50:25,2
422,370
422,371
E
-13765,705
+13526,834
======================================
201:00:10,0
201:00:30,7
376,260
376,274
F
-13900,602
+13175,572
======================================
229:19:40,0
229:20:10,5
309,840
309,893
B
-14135,670
+12973,640
xxiv
SUTP101
CIVIL ENGINEERING I
DY =
DIS =
+0,139
1425
DX =
DS =
-0,001
0,139
CLASS = ?
3.
Traverse Adjustment using the Bowditch Proportional [DY, DX] Method
Adjusted Data Traverse between Points S and T
Observed
Final
Data
Data
Name
Y
X
——————————————————————————————————————————————————————————————————————
S
+13142,910
+6295,840
291:30:10,0
291:30:10,6
248,290
248,294
T1
+12911,897
+6386,852
======================================
303:30:28,0
303:30:28,8
199,490
199,494
T2
+12745,558
+6496,983
======================================
2:30:36,0
2:30:35,6
200,820
200,825
T3
+12754,352
+6697,615
======================================
8:35:04,0
8:35:02,8
227,700
227,705
T
+12788,340
+6922,770
DY =
DIS =
CLASS = ?
xxv
+0,007
876
DX =
DS =
-0,015
0,017
SUTP101
CIVIL ENGINEERING I
Further Exercises (No solutions available)
1.
1
2
3
4
5
6
7
@Cable Hill
Cato 
256.25.20
256.25.42
Phala 
129.19.10
129.19.28
Trav1
6.12.20
@Trav1
Cable Hill
186.12.30
Trav2
18.16.13
@Trav2
Trav1
198.16.40
Hartebeest
34.55.19
@ Hartebeest
Trav2
214.54.27
Island 
126.29.29
126.29.58
Sydenham 
243.55.49
243.56.22
Edenvale 
319.59.59
320.00.33
xxvi
SUTP101
2.
CIVIL ENGINEERING I
A traverse was run between two trig beacons TATE and SALVATION
1
2
3
4
5
6
7
@TATE
HIGH RIDGE
266.15.59
266.16.27
WOODLANDS
13.17.50
13.18.27
W16
37.09.54
SQUEEZ
137.14.50
137.15.15
@W16
TATE
217.09.42
T2
7.26.14
@T2
W16
187.26.13
T1
29.51.10
@T1
T2
209.51.18
SALVATION
25.02.22
@SALVATION
SQUEEZ
148.02.20
148.02.10
T1
205.02.38
MAN HOUSE
281.31.33
281.31.20
REKAJU
45.42.59
45.42.44
xxvii
SUTP101
CIVIL ENGINEERING I
COMPLETE THE TRAVERSE CALCULATION:
Co-ordinates
Y
SALVATION
+3357,49
TATE
+2663,99
X
+ 303801,56
+ 302645,23
REDUCED DISTANCES
TATE - W16 830,885m
T2 - T1 162,649m
W16 - T2 163,355m
T1 - SALVATION
210,823m
3.
1
2
3
4
5
6
7
@Rooikrans
Blouberg
317.15.27
+20
317.15.47
Pafuri
220.29.59
=20
+28
220.30.27
Trav1
121.35.01
-03
121.35.22
25
+24
@Trav1
Rooikrans
301.34.50
Trav2
47.33.55
25
-25
+30
-07
47.33.23
@Trav2
Trav1
227.33.40
+30
-10
Chengeta
198.26.11
+26.01
-10
198. 25.51
Trav2
18.26.18
25.48
+03
18.25.51
Hartley
339.19.56
-30
339.19.26
Chegutu
167.51.11
-31
167.50.40
Selous
222.27.26
-29
222.26.57
@ Chengeta
-30
18.25.48 - 18.26.01 = -13/4 =- 3
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CIVIL ENGINEERING I
4.
1
2
3
4
5
6
7
@ Katskop ∆
Redcliff ∆
231.37.00
231.36.55
Waterloo ∆
356.33.00
356.33.02
H1
201.24.10
@H1
Katskop ∆
21.24.00
H2
225.25.40
@H2
H1
45.25.50
H3
231.37.00
@H3
H2
51.37.00
Redcliff ∆
247.08.18
@ Redcliff ∆
Mt Moriah ∆
312.10.50
312.10.48
Katskop ∆
51.36.50
51.36.55
H3
67.08.12
xxix
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CIVIL ENGINEERING I
Chapter 6
SITE SURVEYING
6.1
Introduction
Site surveys are probably the most common surveys conducted for engineering applications.
Running such surveys requires the same principles applicable to other surveys. Once the
site for construction has been identified, any existing control points in the vicinity must be
identified and all relevant information pertaining to these control points collected from
(topographic) maps and any other associated records.
A reconnaissance of the area must then be conducted to identify suitable points for the
establishment of observation points (stations). The identified points are usually marked with
iron or wooden pegs. A survey is then run to coordinate the observation points.
Coordination of points may require a simple polar survey, traverse, or any other
convenient survey technique depending on the circumstances.
Once observation points have been coordinated, a survey is run to pick the positions
of all features of interest on site. Features to observe include points with noticeable
change of slope, topographical features, and any other features of interest that you
think should not miss out on the map.
Every point to which observations are made must be numbered, and its attribute
data recorded. If recommended that sketches be made on the field sheet to enable
the details to be correctly drawn on the plan. As a rough guide, observations should
be approximately 5m to 20m apart, depending on the terrain.
Both before beginning to take the observations and after taking them, the directions
to known points (Reference Objects (RO)) need to be taken at every observation
station. The direction readings need to be compared and a survey from a particular
station should only be accepted if the closing direction disagrees with the starting
direction by not more than about a minute. Otherwise the instrument must have
evidently slipped or bumped sometime during the observing, and all the
observations taken at that set-up will have to be scrapped and repeated.
6.2 Carrying out a site survey
The phases in carrying out a site survey can be summarised in three phases; field
survey data collection, field survey data processing and presentation of survey data
results.
6.2.1
Field Survey data collection
xxx
SUTP101
CIVIL ENGINEERING I
a).
Identification of Control points

Identify control points in the vicinity of the site
xxxi
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CIVIL ENGINEERING I
b). Orientation

Observe to at least two known points to orient your observations
xxxii
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CIVIL ENGINEERING I
c). Observation shots

Take observation shots to staff / prism held at positions of interest

Take observation shots to staff/prism positioned at any change of gradient
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CIVIL ENGINEERING I
6.2.2 Instruments and observation reading process
a). Automatic Level with Horizontal circle and staves


Setting up, centring and levelling up
•

Setting up and levelling up same as in levelling, but needs centring up over
the station with the help of a plumb bob
Orientation
•
•
•

Set correct direction to first point
Check reading to second point
Record readings
Taking readings to staff
•
•

This is the simplest method
Some procedure as in levelling but also read horizontal circle
Record readings together with attribute of staff position at every point of
interest
Check orientation to first point (RO)
b). Theodolite with staves

This is the tacheometric (tache) method common before the advent of a Total
station.

It is now superseded by the total station
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CIVIL ENGINEERING I
c). Total station

The most popular method in use presently

Setting up, centring and levelling up
•
■
■
■

Orientation
•
•
•

•

Same as in angle measurement
Centring over station mark
level circular bubble
level main (digital) bubble
Set correct direction to first point
Check reading to second point
Record observations
Taking readings to prism
Record observations together with attribute of prism position at every point of
interest
Check orientation to first point (RO)
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CIVIL ENGINEERING I
6.2.3
Field Survey data processing
a). Calculation and analysis of field data

Data collected with Automatic level
•
•
•

You may use the rise and fall method or Height of Instrument method to
calculate heights of points
Check closing error is within limit
Calculate horizontal distances from 100(U-L)
Data collected manually with Total Station
•
Calculate horizontal distances to observed points
Horizontal Distance = S.SinZ where S is the measure slope distance.
•
Calculate point heights
Height of Instrument (HI) +/- Δh – Height of Object (HO = Prism height)
where Δh = S.Cos Z
Z = Vertical reading (not angle)
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CIVIL ENGINEERING I
b). Plotting of points

Manually (by hand)
•
•
Place protractor over station position on plan
Number each point and mark the horizontal direction of each numbered point
on paper at the edge of the protractor
Using a suitable scale, e.g. 1/1000, 1/500; measure out the distance from the
station along the direction to each numbered point
Mark the plotted point with a decimal and write in its height.
Points describing certain features are linked with lines using a rule to create
outlines of objects with the help of the descriptions (attributes) on the sketch
drawn during field work
•
•
•
6.2.4
Presentation of site survey results

Results of a site survey:
•
•
Are usually presented in form of a large-scale map or plan showing
various features and elevation changes.
Before a final plot is made the draft plan need to be checked for:
■
sufficient spot shots
■
suspicious contours
■
data omissions
A field check with the plan-in-hand is the professionally recommended way
of carrying a check.
•
Height information may be presented using any of the products discussed
above; spot-heights or contours on maps or DEM which subsequently can
be used for different form of terrain analysis
6.3 Height information

Height information can be represented in the form of spot heights and contours.
a) Spot-heights
 A spot-height is a written height value.
 the point the height refers to is the decimal comma or point
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CIVIL ENGINEERING I

some large-scale plans may show every measured height used to survey
the area and the spot-heights are then also called spot-shots
b) Contours

A contour line joins points that have the same particular height on the
survey site
 Small contour interval (vertical separation) for site plans recommended;
e.g: 0.5m, 1m, 10m
 The height of a particular contour is written along its length, upward in
the up-slope direction.
• Interpolating contours from spot–heights (spot shots)
▪
Real terrain
▪
Interpolation of contours
There are a number of methods that can be used to interpolate between spot shots.
Done in DRAWING I.
6.4
Summary of contour characteristics
1.
2.
3.
4.
5.
Closely spaced contours indicate steep slopes
Widely spaced contours indicate moderate slopes
Contours must be labelled
Contours are not shown going through buildings
Contour lines cannot cross
xxxviii
SUTP101
6.
7.
8.
CIVIL ENGINEERING I
Depressions and hills look the same – the contour value will distinguish the terrain
The ground slope between contour lines is uniform. If not, it means not enough
spot shots were taken
Contour lines tend to be parallel to each other on uniform slopes
PTO FOR WORKED EXAMPLE OF MEASUREMENTS FOR SITE SURVEY
xxxix
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SURVEYING: THEORY AND PRACTICAL
WORKED EXAMPLE: SITE SURVEY FIELD BOOK AND CALCULATIONS
(Calculated values are shown on BOLD ITALICS)
STATION
From
To
ANGLE/ DIRECTION
Join
Direction
Observed
Horizontal
Direction
Vertical
Reading
(Z)
DISTANCES
Slope
(S)
Hor.
(S.SinZ)
T1
HI / HO
HI - HO
±Δh
(S.CosZ)
1,52
RED
LEVEL:
HI – HO
±Δh
20,50
DESC
16mm
Iron Peg
T2
92.34.00
92.34.10
(orientation)
20mm
Iron Pipe
T3
164.55.20
164.55.27
(orientation)
25mm
Iron Pipe
1
74.35.10
91.34.37
8,23
8,23
1,24
+0,28
-0,23
20,55
BB
2
64.12.31
88.17.11
20,86
20,85
1,50
+0.02
+0,62
21,14
MH
3
84.44.25
93.15.00
12,77
12,75
1,56
-0,04
-0,72
19,74
SR
4
97.22.35
94.33.15
24,22
24,14
1,62
-0,10
-1,92
18,48
TR
5
102.52.54
87.18.00
32,79
32,75
1,16
+0,36
+1,54
22,40
TB
6
125.17.45
88.55.10
45,01
45,00
2,23
-0,71
+0,85
20,64
ELP
T2
(RO)
92.34.15
(orientation)
Date revised: November 2023
SUTP101
CIVIL ENGINEERING I
EXERCISE
STATION
From
To
ANGLE/ DIRECTION
Join
Direction
Observed
Horizontal
Direction
Vertical
Reading
(Z)
DISTANCES
Slope
(S)
Hor.
(S.SinZ)
P1
HI / HO
1,55
HI - HO
±Δh
(S.CosZ)
RED
LEVEL:
HI – HO
±Δh
125,50
DESC
12mm
Iron Peg
TRIG8
112.28.00
112.27.50
(orientation)
12mm
Iron Pipe
TRIG4
284.56.20
284.56.10
(orientation)
10mm
Iron Peg
11
206.15.31
93.47.51
15,18
1,50
TB
12
242.18.22
84.56.55
25,20
1,50
FC
13
255.55.13
94.52.16
41,16
1,50
HSE
14
262.44.16
95.31.57
44,41
1,50
HDGE
15
270.59.59
88.48.26
52,99
1,50
BB
16
277.48.42
86.21.48
55,86
2,10
TP
TRIG8
(RO)
112.27.55
(orientation)
ii
SUTP101
CIVIL ENGINEERING I
iii
SUTP101
CIVIL ENGINEERING I
Chapter 7
AREAS AND VOLUMES
Calculating the area of a parcel of land requires some geometry. Most of the following examples are
for simple shapes. Choose the shape, or combination of shapes that most closely match the area.
Many land areas have a very irregular shape. In this case, you can use the offset method described in
the final example.
7.1
Revision: Shapes and Formulae - Areas
These shapes are defined by the opposite sides being straight, parallel, and of equal length. The area
of all 3 shapes is found by multiplying the length (L) times the width (W).
Formula
Area = L (length) x W (width)
Example
Area = L (length) x W (width)
L = 75 m, W = 25 m
• Area = 75 x 25
• =1875 m2
•
Circle
The area of a circle is found by multiplying the constant pi (  ) times the square of the radius.
Formula
Area =  x r2
•  (pi) = 3.14
• r2 (radius squared) = r x r
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CIVIL ENGINEERING I
Example
Area =  x r2
•r=6m
Area = π x r2
• =  x (6 x 6)
• =  x 36
• = 113,097 m2
•
Triangle
The area of a triangle is found by multiplying the length of the base times the length of the height,
then dividing this result by 2.
Formula
Area = (b x h) / 2
• b = length of base
• h = length of height
Example
Area = (b x h) / 2
• b = 10 m, h = 5 m
Area = (b x h) / 2
• = (10 x 5) / 2
• = 50 / 2
• = 25 m2
•
Trapezoid
The area of a trapezoid is found by first finding the average length of the parallel sides (A + B) / 2,
then multiplying the result by the height (h).
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CIVIL ENGINEERING I
Formula
Area = [(A + B) / 2] x h
Example
Area = [(A + B) / 2] x h
• A = 20 m, B = 10 m, h = 5m
Area = [{A + B) / 2] x h
• = [(20 + 10) / 2] x 5
• = [30 / 2] x 5
• = 15 x 5
• = 75 m2
•
Oval
The area of an oval is found by multiplying the width (W) times the length (L), then multiplying the
result by 0.8
Formula
Area = (W x L) x 0.8
• W = width
• L = length
Example
Area = (W x L) x 0.8
• W = 10 m , L = 20 m
Area = (W x L) x 0.8
• = (10 x 20) x 0.8
• = 200 x 0.8
• = 160 m2
•
Compound Simple Shapes
Many landscape areas can be sub-divided into multiple, simple shapes. In these cases, use the
formulas for the simple shapes and add the results for the total area. See the appropriate
formula in other sections of this chapter.
Revision 6
SUTP101
•
CIVIL ENGINEERING I
Odd Shapes
The method used for irregular shaped areas is called the "offset method". First measure the
length of the longest axis of the area (line AB). This is called the length line. Next, divide the
length line into equal sections, for example 3 m. At each of these points, measure the distance
across the area in a line perpendicular to the length line at each point (lines C to G). These
lines are called offset lines. Finally, add the lengths of all offset lines and multiply the result
times the distance that separates these lines (3m in this example).
Example
Length line (AB) = 18 m., distance between offset lines is 3m apart
• Length of each offset line
C = 15 m, D = 10 mt, E = 15 m, F = 25 m, G = 20 m
Total length of offset lines = C + D + E + F + G
• = 15 + 10 + 15 + 25 + 20
• = 85 m
Area = Distance between offset lines x sum of the length of the offset lines
• = 3m x 85 m
• = 255m2
•
Formulae summary
Shape
Equation
Variables
Regular triangle
(equilateral triangle)
s is the length of one side of the triangle.
Triangle
a and b are any two sides, and C is the angle between
Triangle
b and h are the base and altitude (measured
Square
Rectangle
them.
perpendicular to the base), respectively.
s is the length of one side of the square.
l and w are the lengths of the rectangle's sides (length
and width).
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CIVIL ENGINEERING I
Rhombus
a and b are the lengths of the two diagonals of the
Parallelogram
b and h are the length of the base and the length of the
Trapezoid
a and b are the parallel sides and h the distance
Regular hexagon
s is the length of one side of the hexagon.
Regular octagon
s is the length of one side of the octagon.
Circle
r is the radius and d the diameter.
Circular sector
r and θ are the radius and angle (in radians),
Ellipse
a and b are the semi-major and semi-minor axes,
Total surface area of a
Cylinder
r and h are the radius and height, respectively.
Total surface area of a
Cone
r and l are the radius and slant height, respectively.
Total surface area of a
Sphere
r and d are the radius and diameter, respectively.
7.2
rhombus.
perpendicular height, respectively.
(height) between the parallels.
respectively.
respectively.
Calculation of Areas using Coordinates
When a piece of land is surveyed, it is often required to state the area in Hectares to four decimal
places. This can successfully be done by calculating the area using coordinates of the corner beacons.
Formulae:
Σ(Y x X) = Y1.X2 + Y2.X3 + Y3.X4…….+Y(n-1).Xn + Yn.X1
Σ(X x Y) = X1.Y2 + X2.Y3 + X3.Y4…….+X(n-1).Yn + Xn.Y1
It is usual practise to number the points of irregular figures in a clockwise order.
Example
Calculate the area of figure FGH by coordinates. Give the answer in hectares to four decimal
places.
Revision 8
SUTP101
F
G
H
CIVIL ENGINEERING I
-1023,560
- 842,170
-1122,072
+2174,930
+1886,480
+1893,279
Layout of calculation:
F(1)
G(2)
H(3)
F(1)
Y
-1023,560
- 842,170
-1122,072
-1023,560
X
+2174,930
+1886,480
+1893,279
+2174,930
÷2
÷10 000
YxX
XxY
-1930925,47
-1594462,78
-2440428,06
-1831660,80
-2116766,39
-1937884,65
-5965816,30
-5886311,84
79504,46
39752,23m²
-5886311,84
-5965816,30
79504,46
3,9752ha
Exercises
1.
Calculate the figure SURVE by using coordinates. Give the answer in hectares to four
decimal places.
Revision 9
SUTP101
E
R
S
U
V
CIVIL ENGINEERING I
+1881,790
+ 642,450
+2091,560
+1242,470
+ 440,870
+3028,620
+2536,470
+1544,600
+ 759,950
+2944,060
Answer: Area (ha) = 226,3012
2.
A
B
C
D
Calculate the figure ABCD by using coordinates. Give the answer in hectares to four decimal
places.
+1275,210
+1376,080
+1018,110
+1109,220
+ 957,370
+1301,270
+1120,330 +1407,330
Answer: Area (ha) = 4,3552
7.3
Revision: Shapes and Formulae - Volumes
Revision 10
SUTP101
CIVIL ENGINEERING I
Rectangular Solid
Volume = Length X Width X Height
V = lwh
Surface = 2lw + 2lh + 2wh
Prisms
Volume = Base X Height
v=bh
Surface = 2b + Ph (b is the area of the base P is
the perimeter of the base)
Cylinder
Volume = r2 x height
V = r2 h
Surface = 2 radius x height
S = 2rh + 2r2
Pyramid
V = 1/3 bh
b is the area of the base
Surface Area: Add the area of the base to the sum
of the areas of all of the triangular faces. The areas
of the triangular faces will have different formulas
for different shaped bases.
Cones
Volume = 1/3 r2 x height
V= 1/3 r2h
Surface = r2 + rs
S = r2 + rs
=r2 + r
Sphere
Volume = 4/3 r3
V = 4/3 r3
Surface = 4r2
S = 4r2
7.4 Calculation of volumes from cross-sections
There are three basic methods:
Revision 11
SUTP101
•
CIVIL ENGINEERING I
By mean areas
The volume is determined by multiplying the mean of the cross-sectional areas by the total
distance between the end sections.
V=
A1 + A 2 + A 3 + .......... + A n −1 + A n
xL
n
This is not a very accurate method and is inclined to give volumes which are too large.
•
By end areas (Trapezoidal rule)
Let A1 and A2 be the areas of two cross-sections a distance “d” apart. The volume between
them will be
V=
A1 + A 2
xd
2
provided the cross-section midway between A1 and A2 is approximately the mean of these
two.
If a number of cross-sections A1, A2, A3, A4………… An-1, all a distance “d” apart are considered,
the total volume is
 A + An

V=dx  1
+ A2 + A3 + A4 + .............. + An −1 
 2

Where d usually = 20m
This is the Trapezoidal rule for volumes.
•
By Simpson’s rule (Prismoidal rule)
The two previous methods do not give very accurate results and the best results are achieved
by assuming the volume of earth between two successive cross-sections resemble a prismoid.
A prismoid is a solid made up of two end faces which must be parallel plane figures not
necessarily of the same shape. The volume of a prismoid is given by
V=
x
where A1 and A2 are the end areas a distance d apart (usually 20m) and “m” is the area of the
section midway between. When a large number of cross-sections are used, every alternative
Revision 12
SUTP101
CIVIL ENGINEERING I
section may be regarded as an end-face and the other sections taken as respective midsections (m).
Simpson’s rule for volumes is a development of the Prismoidal rule and is
V=
d
+ A1 + 4 A2 + 2 A3 + 4 A4 + 2 A5 + ............. + 4 An −1 + An
3
Where d is usually = 20m.
NOTE:
8.4.1
For this rule there must be an ODD number of cross-sections!
Calculation of area of cross-section
1/s
1/s
b
d
d1
b   ns 

d = c +  x 
2s   n + s 

b   ns 

d1 = c +  x 
2s   n − s 

Area =
d x d1 b 2
s
4s
Exercises
1.
Calculate the area of the following cross-section:
n = 100
s=2
b = 30m
c = 2m
Revision 13
SUTP101
2.
CIVIL ENGINEERING I
Calculate the area of the following cross-section:
n = 20
3.
s = 1,5
b = 10m
c = 2m
Calculate the volume of material excavated in a road cutting where the following crosssectional areas were obtained. Use the Trapezoidal rule and then Simpson’s rule, and
compare your answers.
Chainage
Area (m²)
0
20
40
60
80
100
120
0
35
15
40
10
20
0
CIVIL ENGINEERING
S2: SURVEYING THEORY AND PRACTICAL
Revision 14
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CIVIL ENGINEERING I
Chapter 1 REVISION:
SURVEYING I
IMPORTANT ADDITIONAL INFORMATION:
ETHEKWINI MUNICIPALITY SURVEY HANDBOOK Sixth Edition (EMSH)
EMSH:
Chapter I (Pages 1 to 36)
Chapter V (Pages 88 to 97)
Chapter VI (Pages 98 to 104)
Chapter VIII (Pages 123 to 130)
From the Surveying I (SURV111) Course Notes – see extracts below:
Chapter / Section
Contents
Page
2.
THE SOUTH AFRICAN COORDINATE SYSTEM, AND BASIC COORDINATE CALCULATIONS
2.3
The South African Coordinate System
14
2.3.1
The Grid and Graticule
15
2.3.2
Angles of Direction
15
2.3.3
Orientation
16
2.4
Calculations (Coordinate) – S.A. Coordinate System
16
2.4.1
Reference Systems
17
2.4.2
The Terms used in Coordinate Calculations
18
2.4.3
The Polar on Electronic Calculator
19
2.4.4
The Join on Electronic Calculator
19
4.
ANGLE AND DISTANCE MEASUREMENT
4.1
Introduction to Electromagnetic Distance Measurement
61
4.2
Geometric Corrections
62
5.
TRAVERSING
5.2
Traverse field work
73
5.2.1
74
Orientation
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CIVIL ENGINEERING I
5.2.2
The observation procedure
74
5.2.3
Reduction of the field book
74
5.3
The traverse direction sheet
75
5.4
The traverse calculation
81
REVISION FROM SURV111 COURSE NOTES:
THE SOUTH AFRICAN COORDINATE SYSTEM, AND BASIC COORDINATE
CALCULATIONS
2.3
The South African Coordinate System
The South African coordinate system is based on the "GAUSS
CONFORM PROJECTION". This system consists of belts running NORTH
and SOUTH, two degrees of longitude wide, with the central meridian being
the odd meridian.
▪
The Y coordinate is measured positive to the west of the origin and negative to the
east. The X coordinate value increases from zero at the equator in a southern direction
and, therefore, always positive in South Africa.
.
In the northern hemisphere and, in some systems, in the southern hemisphere, these
conventions are reversed. In survey work angles are always measured in a clockwise
direction.
Revision 16
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CIVIL ENGINEERING I
▪
▪
▪
▪
▪
▪
The co-ordinate grid is superimposed onto the geographical graticule, the zero Y grid
line being collinear with the central meridian. The meridians converge as we move
from the equator towards the pole, and will be symmetrical about the central meridian
in each belt.
The co-ordinates of a point on the boundary meridian between two adjacent systems,
i.e., the even meridian, will have the same X value, and the Y will have the same
numerical value, but will have opposite signs in the two systems. If a survey falls on
or near the boundary meridian, it is usual to calculate the entire survey in the belt in
which the greater portion of
the area falls. For this purpose the co-ordinates of trig beacons falling within an
overlap area of 15 minutes of longitude on either side of the boundary meridian, are
given on both co-ordinate systems.
It should be noted that the X value always gives the true distance of the point,
measured along the central meridian, from the equator. The Y co-ordinate gives the
distance east or west of the central meridian and it is here that scale distortion is
present. On the map these lines are longer than their actual length on the earth's
surface.
Bearings or Azimuths are true geographical angles of direction and are related to true
north or true south.
Hence the bearing of a line and its direction on the co-ordinate system will only
coincide when the point of measurement is situated on the central meridian. The word
"bearing" should thus not be used when dealing with a rectangular co-ordinate
system.
For short lines, their directions on the map are virtually the same as their bearings on
the spheriod, and as a result, the shapes of small figures are similar to their shapes on
the spheroid although, due to scale distortion, their areas are increased proportionally
to their distances away from the central meridian.
2.3.1 The Grid and Graticule
In most survey work coordinates are used, and before plotting them, a
rectangular "coordinate grid" is constructed.
A GRID is the representation on a map of a system of equally spaced straight
lines, parallel to the Y and X axes of the coordinate system, the exact distance
of each line from its parent axis being known. The grid thus consists of a
system of squares of known dimensions.
Knowing the value (distance from the zero axis) of the grid lines, the position
of a point of known coordinates can be plotted by scaling the correct distance
from two nearby grid lines.
A GRATICULE is the representation of the lines of latitude and longitude on
a map.
2.3.2 Angles of Direction
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The direction, angle of direction, or direction angle of any line on any
coordinate system is the angle, measured in a clockwise sense, between zero
direction of the coordinate system and the line.
The zero direction of the South African system is the direction in which the X
coordinate increases positively, whilst the Y coordinate remains unchanged,
ie. true geographical south at the central meridian of each belt.
A direction is always referred to the zero direction of the coordinate system,
and not to any other line, such as true magnetic north. On small local survey
directions are sometimes referred to any arbitrary line and, in such cases, they
are called "Assumed Directions". A line pivoted at the origin and rotated
clockwise starting from the positive direction of the X axis, will sweep out
lines of direction starting form 0° to 360°.
The first quadrant contains direction angles from 0° to 90°.
The second quadrant contains direction angles from 90° to 180°.
The third quadrant contains direction angles from 180° to 270°.
The fourth quadrant contains direction angles from 270° to 360°.
2.3.3 Orientation
Orientation is the rotation of an angle so as to bring the initial line to the zero
direction (in South Africa this is south).
The coordinate values of the control network over the Republic are supplied,
by the Trigonometrical Survey Office, in metres. The heights of the
Trigonometrical Stations (trig stations) above mean sea level are given to the
tops of all trig beacons in metres, unless otherwise stated in the trig lists.
Mean Sea Level is determined by taking tide observations at various points
around the coast, over a continuous period for many years. From these
observations, a mean height of the sea is determined accurately.
The most accurate method of surveying heights is by precise levelling, and a
network of precise levels connected to the tide gauges has been carried out.
The heights of the trig beacons are connected to the precise benchmarks.
These elevations are determined by observing vertical angles and adjusting the
heights to conform with those of the precise levelling network where possible.
This operation is known as trig levelling. In this way, heights can be carried
over the whole country, although many of the trig stations are great distances
from the nearest line of precise levels.
2.4
Calculations (Coordinate) - South African Coordinate System
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The map projection used is the Transverse Mercator Projection (also known as the
Gauss Conformal). This is a cylindrical projection, and has the advantage that
directions remain true for short lengths as you move from the central meridian.
2.4.1 Reference Systems
GRATICULE is a reference system using lines of longitude and latitude, ie. it
is a geographical system.
GRID - the Transverse Mercator projection has a rectangular grid. The N-S
line agrees at the central meridian only.
DISTANCES - are measured on the graticule and must be corrected to sea
level and allowance made for scale distortion.
DIRECTIONS - note that sometimes the term "direction" and "bearing" are
used interchangeably; in South Africa we use "bearing" related to
geographical systems and "direction" related to the grid system.
TRUE NORTH (T.N.) - the true north at a point is the direction of the North
Pole from that point.
The meridian through any point is a true North-South line. We have shown
that meridians converge towards the poles. Therefore, true North directions
also converge towards the Poles. The T.N. at two points will only be parallel
lines if they are on the Equator or on the same meridian.
It is apparent, therefore, that the T.N. line is not a good direction on which to
base the directions used in a survey, because it is not a parallel direction from
all points in the survey.
GRID NORTH (G.N.) - the central meridian of each system is a T.N. (T.S.)
line, but all grid lines are made parallel to or at right angles to this line and, for
survey purposes, are oriented from the Grid North (South) as it is called, ie. a
line parallel to the central meridian.
The angle of direction or the direction of a line is always referred to as G.S. in
South Africa. Although these directions are fairly commonly referred to as
"bearings", this is a misnomer, as true geographical bearings refer to T.N.
MERIDIAN CONVERGENCE - G.N. coincides with T.N. only at the
central meridian of each zone, e.g. as the 31° meridian passes through Durban,
T.N. and G.N. coincide here.
The difference between G.N. and T.N. (G.S. and T.S.) is of opposite sign on
either side of the central meridian. By drawing a figure, the sign of this
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meridian convergence can easily be determined. A maximum value for
meridian convergence in South Africa is about 30 minutes of arc.
MAGNETIC NORTH (M.N.) - is the direction in which a freely suspended
compass needle points.
2.4.2 The Terms used in Coordinate Calculations
a.
b.
c.
d.
e.
f.
g.
Direction is the angle measured from a reference object in a clockwise
sense.
Oriented direction (direction angle) is the angle measured from grid
south. At the central meridian this coincides with geographical or true
south.
Functional angle - this is to simplify calculations when using
logarithmic or natural tables. The function angle is the direction minus
the preceding right angle, e.g. the functional angle of 123° is 33° and
327° is 47°.
Orientation is the measure of the rotation of an angle so as to bring the
initial line to zero direction, i.e. grid south.
Zenith - an imaginary point vertically above the instrument.
Zenith distance is the angle between the zenith and the point sighted. It
is the vertical angle read on most modern theodolites.
Standard methods of calculation. The civil engineering industry
follows the standard method laid down by the Surveyor General for
cadastral surveys.
2.4.3 The Polar on Electronic Calculator
Use the following format:
A-B
x AB
Dist AB
B
YB
Step 1: ΔY =Dist AB x Sin of Direction AB
ΔX =
Dist AB x Cos of Direction AB
Step 2: YB =
XB =
XA + (±ΔX)
YA + (±ΔY)
2.4.4 The Join on Electronic Calculator
Use the following format:
A–B
x AB
Dist AB
Checked by Polar √
Step 1: Find ΔY and ΔX by subtraction Step 2: Find x
AB:
a. Always obtain a value for Tan−1
∆Y
∆X
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b. Place in proper quadrant by inspecting the signs of ΔY and ΔX.
See first quadrant of coordinate system:
Add 0º if the signs are
Direction AB = x AB = α
+Y
YB
YA
Y = Y-difference
90
A
S
XA
X
Y
B
XB
+X
PTO FOR SECOND QUADRANT
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See diagram of the second quadrant above.
∆Y
α1 = Tan−1
∆X
α = 90o − α1 = 90o − Tan−1
x AB = α + 90o = Tan−1
ΔYΔX
{{+− }}∆∆ XY
+ 90º + 90º
Add 180 º if the signs are
PTO FOR THIRD QUADRANT
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See diagram of the third quadrant above.
Direction AB = x AB = α + 180 º
Add 180 º if the signs are
PTO FOR FOURTH QUADRANT
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See diagram of the fourth quadrant above.
α1 = Tan−1
∆Y
∆X
α = 90o − α1 = 90o − Tan−1
x AB = α + 270o= Tan−1
{{+− }}∆∆ YX
ΔYΔX
+ 90º + 270º
Add 360 º if the signs are
Step 3: Find the distance AB:
AB
=
ΔY
or
sin x AB
AB =
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or AB = ΔY2 + ΔX2
(best to use)
ANGLE AND DISTANCE MEASUREMENT
4.1Introduction to Electromagnetic Distance Measurement (EDM)
The advent of EDM equipment has completely revolutionised all surveying procedures,
resulting in a change of emphasis and techniques. Taping distance, with all its associated
problems, has been rendered obsolete for all base-line measurement. Distance can be
measured easily, quickly and with great accuracy, regardless of terrain conditions. Modern
EDM equipment contains hard-wired algorithms for reducing the slope distance to its
horizontal and vertical equivalent. For most engineering surveys, `total stations' combined
with electronic data loggers are now standard equipment on site. Basic theodolites can be
transformed into total stations by add-on, top-mounted EDM modules. A standard
measurement of distance takes between 1,5 and 3 s. Automatic repeated measurements can
be used to improve reliability in difficult atmospheric conditions. Tracking modes, for the
setting out of distance, repeat the measurement every 0.3 s.
The development of EDM has produced fundamental changes in surveying procedures, e.g.
(1)
(2)
(3)
(4)
(5)
Traversing on a grandiose scale, with much greater control of swing errors, is
a standard procedure.
The inclusion of many more measured distances into triangulation, rendering
classical triangulation obsolete. This results in much greater control of scale
error.
Setting-out and photogrammetric control, over large areas, by polar
coordinates from a single base line.
Offshore position fixing.
Deformation monitoring to sub-millimetre accuracies using high-precision
EDM.
The latest developments in EDM equipment provide plug-in recording modules capable of
recording many thousand blocks of data for direct transfer to the computer. There is
practically no surveying operation which does not utilize the speed, economy, accuracy and
reliability of modern EDM equipment.
EDM instruments may be classified according to the type and wavelength of the
electromagnetic energy generated or according to their operational range. Very often one is a
function of the other. Considering the energy generated, the classification is as follows:
(1)
Infra-red radiation (IR) classifies those instruments most commonly used in
engineering. The accuracies required in distance measurement are such that
the measuring wave cannot be used directly due to its poor propagation
characteristics. The measuring wave is therefore superimposed on the highfrequency waves generated, called carrier waves. The superimposition is
achieved by amplitude, frequency or impulse modulation. In the case of IR
instruments, amplitude modulation is used.
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(2)
4.2
Microwave instruments use radio wavelengths as carriers and therefore require
two instruments, one at each end of the line to be measured, that are capable of
receiving and transmitting the signals. The microwave carrier is always
frequency modulated for measuring purposes. As these instruments do not rely
on light being returned to the master instrument by a reflector, they can be
used day or night in most weather conditions. These instruments are capable
of long ranges up to 25 km and beyond, with typical accuracies of ±10 mm ±5
mm/km.
Geometric Corrections
Slope Correction
When the slope distance has been obtained from an EDM measurement, a slope correction
must be applied to it in order to obtain the equivalent horizontal distance. If the EDM and
theodolite are coaxial as in most total stations and the telescope is tilted and pointed at the
centre of the prism the correct slope distance is obtained no matter how the prism is tilted and
the horizontal distance can be calculated. For an integrated total station, this calculation is
carried out by the instrument's microprocessor and the result displayed automatically.
Height Above Datum Correction
When a survey is to be based on the National coordinate system, the line measured must be
reduced to its equivalent length at mean sea level (MSL). The correction is negative unless a
line below MSL is measured.
Scale Enlargement
A map projection provides a means of representing the curved surface of the Earth on a plane
surface so that coordinate grids can be defined and maps drawn. The relative positions of
points on the grid are altered slightly from their ground positions as a result of using a
projection to account for the curvature of the Earth. Therefore, distances calculated from
National coordinates will not, in some cases, agree with their equivalent measured on site.
To convert measured distances to projection (or grid) distances the measured distance is
converted to its equivalent at MSL and a scale factor is used. The value of the scale factor
varies across the country.
Meteorological Corrections
Using EDM equipment, the measurement of distance is obtained by measuring the time of
propagation of electromagnetic waves through the atmosphere. Whilst the velocity of these
waves in a vacuum is known, its value will be reduced according to the atmospheric
conditions through which the waves travel at the time of measurement.
The value of the correction is affected by the temperature, pressure and water vapour content
of the atmosphere as well as by the wavelength of the transmitted electromagnetic waves. It
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follows from this that measurements of these atmospheric conditions are required at the time
of measurement.
EDM equipment is standardized under certain conditions of temperature and pressure. For
instance, some makes of equipment are standardized at 20°C and 1013.25 mbar of pressure,
whilst others are standardized at 12°C and 1013.25 mbar. Under these respective atmospheric
conditions, the measured distances would not require a velocity correction. It follows that
even on low-order surveys the measurement of temperature and pressure is important.
TRAVERSING
5.2
Traverse field work
•
•
•
•
•
Reconnaissance : always plan the route of a traverse, starting from known point and
finishing on known point (unless not possible)
Follow correct observation procedure
Book all the information including what may seem insignificant, e.g. date.
Use well trained assistants familiar with the routine and are able to set up targets
Avoid short traverse legs
5.2.1 Orientation
Once the theodolite has been set up over the starting point of the traverse – a point
with known coordinates – calculate joins to at least two other visible known points for
orientation. Select the longest ray, which in addition is well defined, as your reference
object (RO), and bisect the RO. Then follow the prescribed procedure for the make/
model of theodolite you are using to set the join direction into the instrument.
Check your orientation by bisecting the other known point which is visible from your
instrument position. Compare this direction with the join direction, which should not
differ by more than a few seconds.
5.2.2 The observation procedure (manual booking)
All the points in a clockwise direction are observed. The results are recorded in the
field book starting with the RO and closing again on the same point. It is usual
practise to do the first round of observation with the instrument in the “Face Left”
position (also known as “Circle Left”).
On completion of the CL observations, transit the telescope by turning it in the
vertical plane and rotate the instrument through 180 degrees. Re-observe the RO,
complete the remainder of the CR observations and book the results starting from the
bottom and working to the top.
5.2.3 Reduction of the field book
Once the observations have been booked, they must be meaned as follows:
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Station
CL
CR
Mean
Corr.
Adj. Direction (not
oriented)
B
70 14 15
250 14 25
70 14 20
0
70 14 20
T1
140 12 38
320 12 46
140 12 42
-02
140 12 40
C
310 18 34
130 18 44
310 18 39
-04
310 18 35
RO
70 14 22
250 14 30
70 14 26
-06
@A
B and C were used for orientation, and T1 is the first unknown traverse point.
FURTHER EXAMPLE
Note –
instrument below has horizontal collimation (line of sight) error.
Four known points (A, B, D and E) were used for orientation. Trav1 is the
first unknown traverse point.
Station
CL
CR
Mean
Corr.
Adj. Dirn
A
80.59.28
261.02.14
81.00.51
0
81.00.51
B
154.41.21
334.44.05
154.42.43
-01
154.42.42
Trav1
203.44.29
23.47.23
203.45.56
-01
203.45.55
E
206.30.30
26.33.16
206.31.53
-02
206.31.51
D
294.03.00
114.05.46
294.04.23
-02
294.04.21
RO
80.59.37
261.02.12
81.00.54
-03
@CABLE HILL
5.3
The traverse direction sheet
Note that the direction sheet makes use of joins to obtain a provisional orientation correction
at the starting and closing points of the traverse. For intermediate stations, the surveyor has to
rely on back orientation (to the previous traverse station), or where s/he could rely on
observations to external control points at the intermediate traverse stations.
In the example below no external control points were observed from the intermediate traverse
stations T1 and T2. At those points the surveyor therefore had to rely on back orientation (to
the previously occupied point) in the field. In this course only examples where back
orientation was used will be covered.
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Step 1:
From the field book enter the station names and the reduced field
observations into columns 1 and 2.
Step 2: At the known points (starting and closing points) enter the join directions in
column 7 and underline. At the starting point, find the correction [7-2]
to be applied to the observed directions. Enter these corrections into
column 6. Find the mean correction using all the rays. Apply this
correction to the remaining rays end enter into column 5. This value
gives the oriented forward (outgoing) direction to the first intermediate
(unknown) traverse station. This oriented outgoing ray will now
become the incoming ray at the first intermediate traverse station.
Step 3: Treat the observations at the traverse stations in sequence. Enter the oriented
incoming ray from the previous station into column 3. Find the
orientation correction [3-2] and enter into column 4. Apply this
correction to the value in column 3 to give an oriented outgoing ray to
the next traverse station and enter into column 5. This oriented
outgoing ray will now become the incoming ray at the next traverse
station. REPEAT THIS PROCEDURE UP UNTIL THE
PENULTIMATE STATION, I.E. THE LAST UNKOWN
TRAVERSE STATION BEFORE THE CLOSING POINT.
Step 4: We now have an oriented outgoing ray from the penultimate station to the
known (closing) point on which the traverse closes. We also have an
oriented outgoing ray from the closing point to the penultimate station,
which we obtained as we did for the starting point in Step 2. The
procedure to be followed at the closing station is therefore identical to
the procedure we followed at the starting point.
The difference between the outgoing and incoming rays at the closing
station is the closing error and will be distributed proportionally at
each station.
PTO FOR STEP-BY-STEP EXAMPLE OF A TRAVERSE DIRECTION
SHEET WHERE ONLY BACK ORIENTATION WAS USED AT THE
UNKNOWN TRAVERSE STATIONS.
STEP 1
1
2
3
@A
B
70 14 20
T1
140 12 40
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4
5
6
7
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C
310 18 35
@ T1
A
320 12 50
T2
165 17 53
@ T2
T1
345 17 37
D
150 00 04
@D
E
06 27 44
F
142 18 28
T2
330 00 20
STEP 2
1
2
3
4
5
6
7
+10
70 14 30
+08
310 18 43
@A
B
70 14 20
T1
140 12 40
C
310 18 35
49
+09
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@ T1
A
320 12 50
T2
165 17 53
@ T2
T1
345 17 37
D
150 00 04
@D
E
06 27 44
+12
06 27 56
F
142 18 28
+18
142 18 46
T2
330 00 20
35
+15
STEP 3
1
2
3
4
5
6
7
+10
70 14 30
+08
310 18 43
@A
B
70 14 20
T1
140 12 40
C
310 18 35
49
+09
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@ T1
A
320 12 50
T2
165 17 53
49
-01
52
@ T2
T1
345 17 37
D
150 00 04
52
+15
19
@D
E
06 27 44
+12
06 27 56
F
142 18 28
+18
142 18 46
T2
330 00 20
35
+15
STEP 4
1
2
3
4
5
6
7
+10
70 14 30
+04
140 12 53
+08
310 18 43
@A
B
70 14 20
T1
140 12 40
C
310 18 35
49
+09
@ T1
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A
320 12 50
T2
165 17 53
49
-01
52
+08
165 18 00
19
+12
150 00 31
@ T2
T1
345 17 37
D
150 00 04
52
+15
@D
E
06 27 44
+12
06 27 56
F
142 18 28
+18
142 18 46
T2
330 00 20
-04
330 00 31
35
+15
Closure = (330).00.35 – (150).00.19 = +16
Correction = +16/4 = +04” per set-up
At the closing station the sign of the correction at the first set-up
can be reversed and entered at the closing
station, in this case -04”
5.4
The Traverse Calculation
Use the adjusted directions from the direction sheet, and reduced distances to calculate the
coordinates of the intermediate traverse stations. This is accomplished by doing successive
Polars, writing down only the
The sum of
♠Y and ♠X values on the traverse calculation sheet.
♠Y and ♠X should agree with the Y and X differences between the starting and
closing points, in this case between A and D. Various ways are used to distribute the closing
error but we will use the Bowditch Rule, which states:
Y closing error
Correction to Y difference =
x LENGTH OF LEG
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Length of Traverse
X closing error
Correction to X difference =
x LENGTH OF LEG
Length of Traverse
Linear accuracy = (Yclosing error)2 + (X closing error)2
FROM THE REGULATIONS PROMULGATED IN TERMS OF SECTION 10 OF THE
LAND SURVEY ACT, 1997 (ACT No. 8 OF 1997)
Regulation 5
(a) when the position of a point is determined by polars, traverse, triangulation, trilateration,
GPS or a combination of these methods, the displacement between any observed ray,
measured distance or GPS vector and the equivalent quantity derived from the final
co-ordinates of the point fixed shall not exceed- for Class A : A metres; for Class B :
1,5A metres; for Class C : 3A metres;
where A is equal to-
and S is the distance between the known and the unknown point……………….. provided
further that in the case of a traverse the comparison is made to the misclosure of the traverse,
where S is the total length of the traverse in metres;
Traverse Calculation
Dir/ Dist
Join
♠Y
♠X
(students)
140 12 53
+130,740
-157,002
204,31
+0,040
+0,053
165 18 00
+44,299
-168,856
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Station
Y
X
A
-7345,27
+3927,58
T1
-7214,49
+3770,63
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174,57
+0,034
+0,045
150 00 31
+91,921
-159,267
183,89
+0,036
+0,047
S=562,77
+266,960
-485,125
+0,110
+0,145
T2
-7170,16
+3601,82
D
-7078,20
+3442,60
+267,07
-484,98
Linear closing error = 0,182m. Class A = 0,059m. Closure is below Class C.
PTO FOR REVISION EXERCISES
REVISION EXERCISES
1.
Calculate checked coordinates for point B from the data below. Your answer should
be, within a few mm’s, the same for both polars.
A
+ 571,436
-144,770
Dir AB
103º55'26"
Dist AB
362,498m
D
Dir DB
+842,342
-620,777
11º45'38" Dist DB
397,114m
Solution:
NOTE: FOR JOIN AND POLAR CALCULATIONS ONLY THE TEXT PRINTED IN
BOLD ITALICS BELOW NEEDS TO BE SHOWN
1.
POLAR
=====
From
====
A
A
To
==
+571.436
B
Direction
=========
-144.770
103.55.26
Distance
========
Y
=
X
=
362.498
+923.282
-231.999
From
====
D
D
To
==
Direction
=========
Distance
========
B
11.45.38
397.114
Y
=
+842.342
+923.283
X
=
-620.777
-231.999
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2.
Calculate the reduced distance and the oriented direction from A to E. Your answer
must be fully checked.
A
E
+ 4932,566
+ 3961,301
+ 7892,023
+ 8371,822
Solution:
2.
JOIN
====
From
====
A
A
To
==
+4932.566
E
3.
Give, in point form, a brief description of the South African Coordinate System.
Support your answer with suitable sketches.
4.
Coordinates:
A
B
C
D
Direction
Distance
Y
=========
========
=
+7892.023
296.17.21
1083.311
+3961.301
CHECKED BY POLAR
y
+ 324,413
+ 553,813
+ 746,317
+1033,285
4.1
Calculate joins AB, AC and AD.
5.
Coordinates:
+ 468,229
+ 869,194
+ 399,115
+ 819,726
x
A
+ 296,443
+ 486,326
B
C + 774,494 + 548,323 D E + 155,202 + 875,042
5.1
Calculate joins AB, AC, AD and AE.
5.2
Direction
=
+8371.822
x
y
BF
X
=
209 : 52 : 33 Distance
y
x
+ 566,923
+1044,349
+ 842,899
+ 709,904
BF
=
Calculate unchecked coordinates for F.
6.
Calculate the reduced distance and the oriented direction from J TO K:
J
K
- 5247.331
- 7244.316
+8422.725
+8111,724
Your answer must be fully checked.
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7.
Calculate checked coordinates for point L from the data below. Your answer should
be, within a few mm’s, the same for both polars.
R + 240,429 +8012,395 Dir RL 182º 06'12" Dist RL
335,705m
Z
Dir ZL
8.
+229,481
186º 52'01"
+ 261,324
128º 26'28"
DUKE
+927,401
Dir DUKE – SHACK
-554,328
170º 44'52" Dist DUKE – SHACK
451,546m
+ 1682,47
+1919,52
+ 3333,10 BON ACC +
+ 9273,81 JUMI
+
Calculate the reduced distance and the oriented direction from DURR to KP:
+ 5527,45
+ 2651,98
+ 1891,54
+10 026,87
Calculate the reduced distance and the oriented direction from PENN to IFAFA:
PENN
IFAFA
14.
943,095m
Calculate the reduced distance and the oriented direction from CABLE to JUMI:
DURR
KP
13.
Dist BORDER – SHACK
Calculate the reduced distance and the oriented direction from CABLE to BON ACC:
RES8
+ 1827,45
5526,31 +1124,69
12.
-413,668
Dir BORDER – SHACK
CABLE
5244,68
11.
+ 2744,217 + 5621,498
+ 1821,301 +8266,613
Calculate checked coordinates for point SHACK from the data below. Your answer
should be, within a few mm’s, the same for both polars.
BORDER
10.
11,491m
Calculate the reduced distance and the oriented direction from DRUM to POLE:
DRUM
POLE
9.
+7688,324
Dist ZL
+ 19 261.54 + 5261.30
+ 16 609,18 +9111,11
Calculate the reduced distance and the oriented direction from MHLOTI to UM-R:
Revision 37
SUTP101
CIVIL ENGINEERING I
MHLOTI
UM-R
+ 1818,46
+ 2424,92
+ 19 999,99
+26 060,46
PTO FOR SOLUTIONS
Revision 38
SURV221
SURVEYING II
Solutions
4.
JOINS: ALL CHECKED BY POLAR
=====
From
====
A
A
C
A
A
To
==
B
99.18.12
D
Direction
=========
Distance
========
29.46.28
427.527
63.37.31
461.949
+746.317
791.233
Y
=
+324.413
+553.813
+399.115
+1033.285
X
=
+468.229
+869.194
+819.726
5.
JOIN
====
From
====
A
A
NOTE: STUDENTS TO DO JOINS
To
==
Direction
=========
Distance
========
E
340.01.53
413.581
Y
X
=
=
+296.443
+486.326
+155.202
+875.042
CHECKED BY POLAR
AB, AC and AD and compare answers
POLAR (unchecked)
=====
From
====
B
B
To
==
Direction
=========
Distance
========
F
209.52.33
312.522
Revision 39
Y
=
+566.923
+411.249
X
=
+842.899
+571.909
SURV221
SURVEYING II
JOIN
====
6.
From
====
J
J
To
==
Direction
=========
Distance
========
K
261.08.53
2021.057
Y
X
=
=
-5247.331
+8422.725
-7244.316
+8111.724
CHECKED BY POLAR
7.
POLAR
=====
From
====
R
R
To
==
Direction
=========
Distance
========
L
182.06.12
335.705
From
====
Z
Z
To
==
Direction
=========
Distance
========
L
186.52.01
JOIN
====
11.491
From
====
DRUM
POLE
To
==
Direction
=========
Distance
========
2801.501
+1821.301
8.
DRUM
340.45.56
Y
=
+240.429
+228.108
X
=
+8012.395
+7676.916
Y
=
+229.481
+228.107
X
=
+7688.324
+7676.915
Y
=
+2744.217
+8266.613
X
=
+5621.498
CHECKED BY POLAR
9.
POLAR
=====
Revision 40
SURV221
SURVEYING II
From
====
BORDER
BORDER
To
==
Direction
=========
Distance
========
SHACK
128.2628
943.095
From
====
DUKE
DUKE
To
==
Direction
=========
Distance
========
SHACK
170.4452
451.546
Y
=
+261.324
+1000.001
X
=
-413.668
-1000.000
Y
=
+927.401
+1000.001
X
=
-554.328
-999.999
JOINS: ALL CHECKED BY POLAR
=====
10 - 14
From
====
CABLE
CABLE
To
==
Direction
=========
Distance
========
3832.434
Y
=
+1682.470
+5244.680
X
=
+3333.100
+1919.520
BONACC
111.38.40
RES8
RES8
JUMI
155.35.13
8949.286
+1827.450
+5526.310
+9273.810
+1124.690
DURR
DURR
KP
340.32.02
8628.553
+5527.450
+2651.980
+1891.540
+10026.870
PENN
PENN
IFAFA
325.26.05
4675.046
+19261.540
+16609.180
+5261.300
+9111.110
MHLOTI
MHLOTI
UM-R
05.42.52
6090.738
+1818.460
+2424.92
+19999.990
+26060.46
Revision 41
SURV221
SURVEYING II
FURTHER REVISION EXERCISES
1.
Obtain oriented directions for the calculation of final co ordinates for T1 and T2 from
the information given. Do not calculate the traverse.
1
2
3
4
5
6
7
@ Begin
Berg
196. 03. 07
196.03.14
Blou
83. 36. 56
83.37.04
T1
159. 21. 30
@ T1
T2
110. 20. 26
Begin
339. 21.18
@ T2
End
139. 01. 58
T1
290. 19. 54
@ End
Berg
201. 55. 23
201.55.43
Blou
79. 20. 23
79.20.43
Begin
317. 08. 24
317.08.44
T2
319. 02. 20
Revision 42
SURV221
2.
SURVEYING II
Calculate co-ordinates for D, E and F by a traverse calculation with the following
reduced data:
Oriented Directions
Reduced Distances
AD
DE
EF
FB
316,29m
422,37m
376,26m
309,84m
169.10.30
184.50.20
201.00.10
229.19.40
Co-ordinates
3.
A
B
-13 789,45
-14 135,67
+14 258,36
+12 973,64
Calculate co-ordinates for T1, T2 and T3 by a traverse calculation with the following
reduced data:
Oriented Directions
S – T1
T1 – T2
T2 – T3
T3 – T
Reduced Distances
291.30.10
303.30.28
02.30.36
08.35.04
Co-ordinates
S
T
248,29m
199,49m
200,82m
227,70m
+13 142,91
+12788,34
+6295,84
+6922,77
SOLUTIONS FOR NO’S 2 AND 3 (COMPUTER PRINTOUT)
Traverse Adjustment using the Bowditch Proportional [DY, DX] Method
Adjusted Data Traverse between Points A and B
2.
Observed
Final
Data
Data
Name
Y
X
——————————————————————————————————————————————————————————————————————
A
-13789,450
+14258,360
169:10:30,0
169:10:41,3
316,290
316,286
D
-13730,065
+13947,699
======================================
184:50:20,0
184:50:25,2
422,370
422,371
E
-13765,705
+13526,834
======================================
201:00:10,0
201:00:30,7
376,260
376,274
F
-13900,602
+13175,572
======================================
229:19:40,0
229:20:10,5
309,840
309,893
B
-14135,670
+12973,640
DY =
DIS =
3.
1425
DS =
+0,139
DX =
-0,001
0,139 CLASS = ?
Traverse Adjustment using the Bowditch Proportional [DY, DX] Method
Adjusted Data Traverse between Points S and T
Observed
Final
Revision 43
SURV221
SURVEYING II
Data
Data
Name
Y
X
——————————————————————————————————————————————————————————————————————
S
+13142,910
+6295,840
291:30:10,0
291:30:10,6
248,290
248,294
T1
+12911,897
+6386,852
======================================
303:30:28,0
303:30:28,8
199,490
199,494
T2
+12745,558
+6496,983
======================================
2:30:36,0
2:30:35,6
200,820
200,825
T3
+12754,352
+6697,615
======================================
8:35:04,0
8:35:02,8
227,700
227,705
T
+12788,340
+6922,770
DY =
DIS =
876
DS =
0,017 CLASS = ?
Revision 44
+0,007
DX =
-0,015
SURV221
SURVEYING II
Further Exercises (No solutions available) 1.
1
2
3
4
5
6
7
@Cable Hill
Cato ∆
256.25.20
256.25.42
Phala ∆
129.19.10
129.19.28
Trav1
6.12.20
@Trav1
Cable Hill
186.12.30
Trav2
18.16.13
@Trav2
Trav1
198.16.40
Hartebeest
34.55.19
@ Hartebeest
Trav2
214.54.27
Island ∆
126.29.29
126.29.58
Sydenham ∆
243.55.49
243.56.22
Edenvale ∆
319.59.59
320.00.33
Revision 45
SURV221
2.
SURVEYING II
A traverse was run between two trig beacons TATE and SALVATION
1
2
3
4
5
6
7
@TATE
HIGH RIDGE
266.15.59
266.16.27
WOODLANDS
13.17.50
13.18.27
W16
37.09.54
SQUEEZ
137.14.50
137.15.15
@W16
TATE
217.09.42
T2
7.26.14
@T2
W16
187.26.13
T1
29.51.10
@T1
T2
209.51.18
SALVATION
25.02.22
Revision 46
SURV221
SURVEYING II
@SALVATION
SQUEEZ
148.02.20
148.02.10
T1
205.02.38
MAN HOUSE
281.31.33
281.31.20
REKAJU
45.42.59
45.42.44
COMPLETE THE TRAVERSE CALCULATION:
Co-ordinates
Y
X
SALVATION
+3357,49
TATE
+2663,99
REDUCED DISTANCES
TATE - W16 830,885m
T2 - T1 162,649m
+ 303801,56
+ 302645,23
W16 - T2 163,355m
T1 - SALVATION
210,823m
3.
1
2
3
4
5
6
7
@Rooikrans
Blouberg
317.15.27
317.15.47
Pafuri
220.29.59
220.30.27
Trav1
121.35.01
@Trav1
Rooikrans
301.34.50
Trav2
47.33.55
Revision 47
SURV221
SURVEYING II
@Trav2
Trav1
227.33.40
Chengeta
198.26.11
@ Chengeta
Trav2
18.26.18
Hartley
339.19.56
339.19.26
Chegutu
167.51.11
167.50.40
Selous
222.27.26
222.26.57
4.
1
2
3
4
5
6
7
@ Katskop
Redcliff
231.37.00
231.36.55
Waterloo
356.33.00
356.33.02
H1
201.24.10
@H1
Katskop
21.24.00
H2
225.25.40
Revision 48
SURV221
SURVEYING II
@H2
H1
45.25.50
H3
231.37.00
@H3
H2
51.37.00
Redcliff
247.08.18
@ Redcliff
Mt Moriah
312.10.50
312.10.48
Katskop
51.36.50
51.36.55
H3
67.08.12
Revision 49
SSUR221
SURVEYING II
Chapter 2
CIRCULAR CURVES
EMSH:
Chapter XI (Pages 155 to 174)
A simple circular curve consists of a single arc connecting two tangents. A curve can be
described in terms of its radius, or in terms of the angle subtended at the centre by a chord of
particular length. The standard chord in road and rail works is 20m.
PI
I
ECC
BCC
I/2
I/2
PI = POINT OF INTERSECTION
I = INTERSECTION ANGLE
BCC = BEGINNING CIRCULAR CURVE
ECC = END CIRCULAR CURVE
C = CROWN POINT
LENGTH BCC to PI = ECC to PI
= TANGENT LENGTH (T)
TOTAL LENGTH ALONG THE CURVE FROM BCC to ECC = A
BCC to C = ECC to C = A/2 (ALONG THE CURVE)
CD = CROWN DISTANCE = C to PI
Circular Curves 35
SSUR221
SURVEYING II
EXAMPLE: CALCULATION OF SETTING OUT DATA
Calculate and set out, for use in the field, staking data for a horizontal curve from the
following:
R = 760,00m
Peg interval = 20m
Solution
T = R. Tan

= 83,23m
2
I = 12.30.00
Curve is to the right
A = RI (I in radians) =
Ch BCC = 2524, 60m
xBCC – PI = 00.00.00
R
m
180
= 165,81m
Ch ECC = Ch BCC + A = 2690, 41m
Arc
Offset angle θ
15,40
00.34.50
20
00.45.14
10,41
00.23.33
Field Data
Chainage
Offset
Angle
θ
Deflection
Angle 
@BCC
2524,60
InstrumentD
irection
from BCC
Deflection
Distance
2 R Sin
Arc Length
peg to peg
Chord peg
to peg
2 R Sin
00.00.00
to PI
2540
00.34.50
00.34.50
00.34.50
15,40
15,40
15,40
2560
00.45.14
01.20.04
01.20.04
35,40
20
20
2580
00.45.14
02.05.18
02.05.18
55,40
20
20
2600
00.45.14
02.50.32
02.50.32
75,37
20
20
2620
00.45.14
03.35.46
03.35.46
95,34
20
20
2640
00.45.14
04.21.00
04.21.00
115,29
20
20
2660
00.45.14
05.06.14
05.06.14
135,22
20
20
2680
00.45.14
05.51.28
05.51.28
155,13
20
20
ECC
2690,41
00.23.33
06.15.01
06.15.01
165,48
10,41
10,41
Check
 = 165,81
=A

2
Circular Curves 36
SSUR221
SURVEYING II
Alternatively, the staking data could be calculated from the BCC to the Crown Point, and
then from the ECC to the Crown Point. Ch C = Ch BCC + A/2 = 2607, 50m
Chainage
Arc
Offset angle θ
Arc
Offset angle θ
15,40
00.34.50
7,50
00.16.58
20
00.45.14
12,50
00.28.16
10,41
00.23.33
Offset
Angle
θ
Deflection
Angle 
@BCC
2524,60
InstrumentD
irection
from BCC
Deflection
Distance
2 R Sin
Arc Length
peg to peg
Chord peg
to peg
2 R Sin
00.00.00
to PI
2540
00.34.50
00.34.50
00.34.50
15,40
15,40
15,40
2560
00.45.14
01.20.04
01.20.04
35,40
20
20
2580
00.45.14
02.05.18
02.05.18
55,40
20
20
2600
00.45.14
02.50.32
02.50.32
75,37
20
20
C
2607,50
00.16.58
03.07.30
03.07.30
82,86
7,50
7,50
Check
Chainage
Offset
Angle
θ
 = 82,90 =
A/2

4
Deflection
Angle 
@ECC
2690,41
InstrumentD
irection
from ECC
Deflection
Distance
2 R Sin
Arc Length
peg to peg
Chord peg
to peg
2 R Sin
192.30.00
to PI
2680
00.23.33
00.23.33
192.06.27
10,41
10,41
10,41
2660
00.45.14
01.08.47
191.21.13
30.41
20
20
2640
00.45.14
01.54.01
190.35.59
50,40
20
20
2620
00.45.14
02.39.15
189.50.45
70,39
20
20
C
2607,50
00.28.16
03.07.31
189.22.29
82,87
12,50
12,50
Check
 = 82,91 =
A/2

4
Circular Curves 37
SSUR221
SURVEYING II
EXAMPLE 2
Two road straights intersect at an angle of 10.26.50. The straights are to be connected by a
circular arc to the right of radius 1200m. The chainage of the PI is 2517,30m. Calculate:
(i)
(ii)
(iii)
(iv)
(v)
The Crown Distance
The Tangent Lengths
The Total Length of the curve
The Chainages of the BCC, Crown Point and ECC
Tabulated data for setting out the curve from the BCC to the Crown Point.
Pegs are to be placed at continuous 20-metre chainages
(i)
The Crown Distance
Crown Distance = R(Sec I/2 - 1) = 1200(Sec 10.26.50/2 -1) = 5,00m
(ii)
The Tangent Lengths
T = R tan I/2 = 1200 tan 10.26.50/2 = 109,71m
(iii)
The Total Length of the curve
A = (π.R.I)/180 = 218,81m
(iv)
The Chainages of the BCC, Crown Point and ECC
Ch PI
-T
Ch BCC
+A/2
Ch Crown Pt
+A/2
Ch ECC
(v)
=
=
=
=
2517,30m
- 109,71m
2407,59m
+109,405m
2517,00m
+109,405m
2626,40m
Tabulated data for setting out the curve from the BCC to the Crown Point.
Pegs are to be placed at continuous 20-metre chainages
Circular Curves 38
SSUR221
SURVEYING II
EXERCISES
QUESTION 1
Two road straights intersect at an angle (  ) of 10.20.50. The straights are to be connected by
a circular arc to the right of radius 1220m. The chainage of the BCC is 2441,65m. Pegs are to
be placed at continuous 20-metre chainages
Calculate
1.
2.
3.
4.
5.
The Crown Distance
The Tangent Lengths
The Total Length of the curve
The Chainages of the Crown Point and ECC
Tabulated data for setting out the curve from the BCC to the Crown Point.
Pegs are to be placed at continuous 20-metre chainages
QUESTION 2
•
Calculate setting out data for each curve below, and tabulate the data in the standard
format required for setting out in the field. Do all calculation checks.
•
Peg interval: 20,00m on continuous chainages. All the curves are to the right. Then
calculate all the curves assuming they are to the left.
Curve data:
•
R = 80m
I = 120º
Ch. BCC = 36,10m
Direction PI to the BCC is 42.26.51.
•
R = 85m
I = 124º
Ch. BCC = 126,20m
Direction PI to the BCC is 52.20.21.
•
R = 89m
I = 128º
Ch. PI = 306,30m
Direction PI to the BCC is 58.27.45.
•
R = 79m
I = 119º
Ch. PI = 620,40m
Direction PI to the BCC is 138.27.22
•
R = 74m
I = 116º
Ch. PI = 564,50m
Direction PI to the BCC is 244.24.18
Circular Curves 39
SSUR221
SURVEYING II
SUMMARY OF FORMULAE: CIRCULAR CURVE
A
R

T
θ

a
c
=
=
=
=
=
=
=
=
Total Length of Curve from BCC to ECC
Radius of Curve
Intersection Angle of two straights
Tangent Length
Offset angle
θ = Deflection angle
Arc Length between points on the curve (i.e. a portion of A)
Chord length
a
28,648a
1718,87a
1718 , 87a
radians =
deg =
min =
deg
2R
R
R
60R
1.

=
2.
T
=
3.
A
= RI (I in radians) =
4.
Crown Distance (CD) =
5.
c
=
R. Tan

2
R
m
180
R (Sec

- 1)
2
2 R Sin
Circular Curves 40
SSUR221
SURVEYING II
Chapter 3
TRANSITION CURVES
EMSH:
Chapter XI (Pages 175 to 180)
The principal advantages of a transition curve on a horizontal alignment are:
•
A properly designed transition curve provides a natural easy-to-follow path for
drivers, such that the centrifugal force increases and decreases gradually as the
vehicle enters and leaves the circular curve. This minimises the encroachment upon
adjoining traffic lanes, tends to promote uniformity of speed and results in increased
safety.
•
The transition curve length provides a convenient desirable arrangement for
superelevation runoff. Where superelevation runoff is effected without a transition
curve, usually partly on the tangent and partly on the circular curve, the driver
approaching the curve may have to steer opposite to the direction of the curve when
on the superelevated tangent portion in order to keep the vehicle on the straight. This
is an unnatural manoeuvre and explains in part why many vehicles drift to the inside
of the curve.
•
The appearance of a highway is enhanced by the application of a spiral. Their use
avoids the noticeable breaks at the beginning and end of a circular curve, which may
be distorted further by superelevation runoff. Spirals are essential parts of a natural
flowing alignment.
Requirements of a transition curve
•
It must be tangential to the straight. At the BTC R =  and the curvature is 0.
•
The curvature at the end of the transition curve must equal the curvature of the
circular curve.
•
The curvature along the transition curve must increase uniformly.
•
Curvature should increase with superelevation
•
The full super elevation must be achieved at the junction of the circular curve.
Field procedure
1.
Set up at the BTC and stake the spiral up to the BCC. This is done by setting up over
the BTC and sighting the PI, or if the BTC had been placed from coordinates, then by
sighting external control.
2.
Lay off successive deflection angles until the spiral is pegged up to the BCC. If
possible, the BTC, BCC, Crown Point, ECC and ETC should be placed before the
staking of the curves commence, preferable from coordinates.
Transition Curves 41
SSUR221
3.
SURVEYING II
After staking the first spiral, it can be checked by measuring the offset off the tangent
to the BTC and comparing it with
L2
6R
The spiral can be staked by measuring successive chord lengths from peg to peg. For
greatest accuracy the chord length for an arc of length  is calculated by first
computing the relevant values of y and x and then calculating the equivalent chord
length from
y=
4.
(x 2 − x 1 ) 2 + (y 2 − y 1 ) 2
5.
It will usually not be necessary to calculate the chord lengths since they will be
virtually the same as the corresponding arc lengths. Treat each case on its own merits
by doing a quick check between the last two full chainage pegs on the spiral which is
where the curvature is the greatest.
6.
The spiral can also be pegged by taping to every peg from the BTC. In this case the
value of every chord from the BTC to every peg on the spiral must be calculated.
7.
Set up at the ETC and stake the spiral up to the ECC, using the same procedure as
before.
8.
Set up at the BCC and/ or ECC and set out the circular curve. Any closing error
should be adjusted on the circular curve, never on the spiral.
WORKED EXAMPLES
QUESTION 1
The following data refer to a curve system to the right consisting of two Euler spiral
transition curves with a portion of a circular curve between them:
Deflection angle (I) = 50
Lengths of spirals = 100m each
Radius (R) = 320m
Chainage of BTC = 416,00m
Transition Curves 42
SSUR221
SURVEYING II
PI
BCC
I
ECC
BTC
ETC
Calculate:
(1)
(2)
(3)
(4)
(5)
The Apex Distance
The Total Length of the curve system
The Chainages of the BCC, Crown Point, ECC and ETC
Staking data to peg the first transition from BTC to BCC.
Assume the direction BTC - PI to be 00.00.00
The reading that must be observed from BCC to BTC
to give a ZERO reading along the tangent at BCC
Transition Curves 43
SSUR221
SURVEYING II
SOLUTION
(1)
The Apex Distance
PI
SH
BCC
BTC
Shift SH =
L2
L4
L6
−
+
24 R 2688R 3 506880R 5
=
Transition Curves 44
1,301m
SSUR221
SURVEYING II
Ι
L
L3
L5
Apex Distance AD = (R + SH) Tan
=
+
+
2
2 240R 2 34560R 4
(2)
The Total Length of the curve system
S =
S2
rad
2 RL
Total length = 2L +
(3)
L =
L
rad
2R
=
1718 , 87 L
deg
60R
πR(Ι - 2φL )
where  L = 08.57.09 = 379,252m
180
The Chainages of the BCC, Crown Point, ECC and ETC
Ch BTC
+L
Ch BCC
+1/2 cptn
Ch CP
+1/2 cptn
Ch ECC
+L
Ch ETC
=
=
=
=
=
=
=
=
=
416,00
+100,00
516,00
+ 89,626
605,626
+ 89,626
695,252
+100,00
795,252
Transition Curves 45
199,784m
SSUR221
(4)
SURVEYING II
Staking data to peg the first transition from BTC to BCC.
Assume the direction BTC - PI to be 00.00.00
S =
S
3
=
S2
572, 958 S 2
rad =
deg
6 RL
60 RL
S5
S9
Chord lengths from BTC / ETC on the spiral = d = S −
+
90 L2 R 2 22680L4 R 4
S
Chainage
S
416,00
@BTC
420
04
04
00.00.17
00.00.17
440
24
24
00.10.19
00.10.19
460
44
43,998
00.34.40
00.34.40
480
64
63,988
01.13.21
01.13.21
500
84
83,955
02.06.21
02.06.21
516,00(BCC)
100
99,891
02.59.04
02.59.04
Check
d
Inst Dir
00.00.00 to PI
L
= 08.57.09 /3 =  L = 02.59.03 OK
3
Transition Curves 46
SSUR221
(5)
SURVEYING II
The reading that must be observed from BCC to BTC
to give a ZERO reading along the tangent at BCC
Set instrument @ direction 180 -
2 L
= 180 - 05.58.06 = 174.01.54
3
QUESTION 2
Two straights are to be joined by two spiral transition curves each 120m long, plus a portion
of a circular curve of radius 250m. The curve system is to the right. The deflection angle is
62.20.10. Calculate:
(1)
The Apex Distance
(2)
Setting out data to peg the curve from the beginning of the curve (BTC) up to the
beginning of the circular portion (BCC), at 20m continuous intervals. Assume the
direction from the BTC to the PI to be zero. The chainage of the BTC is 169,63m.
(3)
Calculate data to orient the instrument at the BCC so that you would be able to set out
the portion from the BCC to the Crown Point from a zero instrument direction.
(4)
Calculate setting out data to set out the portion from the BCC to the Crown Point at
20m continuous intervals.
Transition Curves 47
SSUR221
SURVEYING II
SOLUTION
1. The Apex Distance
Shift SH = 2,40m
AD = 212,552m
2. Setting out data to peg the curve from the beginning of the curve (BTC) up to the
beginning of the circular portion (BCC), at 20m continuous intervals.
 L = 13.45.04
POINT
CHAINAGE
S
d
 S = DIR
BTC
169,63
0
Students to
attempt
00.00.00 TO PI
180
10,37
00.02.03
200
30,37
00.17.37
220
50,37
00.48.28
240
70,37
01.34.35
260
90,37
02.35.59
280
110,37
03.52.40
289,63
120
04.35.02
BCC
Check
L
= 04.35.01 = OK
3
3. Calculate data to orient the instrument at the BCC so that you would be able to set out
the portion from the BCC to the crown point from a zero instrument direction.
Orientation at BCC: Direction BCC - BTC = 180 -
2 L
= 170.49.57
3
4. Calculate setting out data to set out the portion from the BCC to the Crown Point at
20m continuous intervals.
Length of circular portion =
πR(Ι - 2φL )
= 151,99m
180
Chainage of Crown Point = 289,63 + 151,99/2 = 365,63m

=
a
28,648
1718,87a
1718 , 87a
radians =
deg =
min =
deg
2R
R
R
60R
Transition Curves 48
SSUR221
SURVEYING II
Chainage
Offset
Angle
θ
Deflection
Angle 
InstrumentD
irection at
BCC
@BCC
289,63
0
00.00.00
00.00.00
300
01.11.18
01.11.18
01.11.18
10,37
320
02.17.31
03.28.49
03.28.49
20
340
02.17.31
05.46.20
05.46.20
20
360
02.17.31
08.03.51
08.03.51
20
C
365,63
00.38.43
08.42.34
08.42.34
5,63
Check:
Deflection
Distance
2 R Sin
 − 2 L
= 08.42.31 = OK
4
Arc Length
peg to peg
Chord peg
to peg
2 R Sin
 = 76,00 =
A/2
Do join if using co-ords
QUESTION 3
The data below refers to a curve system to the right consisting of two spirals and a circular
portion.
Deflection angle (I)
Lengths of spirals
= 52
= 100m each
Radius (R) = 300m
Chainage of BTC = 226,00m
Calculate:
(1)
(2)
(3)
(4)
The Total Length of the curve system
The Chainages of the BCC, Crown Point, ECC and ETC
Staking data to peg the first transition from BTC to BCC.
Assume the direction BTC - PI to be 00.00.00
The reading that must be observed from BCC to BTC to give a ZERO reading along
the tangent at BCC
QUESTION 4
Two straights are to be joined by two spiral transition curves each 120m long, plus a portion of a
circular curve of radius 200m. The curve system is to the right. The deflection angle is 61.20.10.
Calculate:
(1)
(2)
(3)
(4)
The tangent lengths
Setting out data to peg the curve from the beginning of the curve (BTC) up to the beginning
of the circular portion (BCC), at 20m continuous intervals. Assume the direction from the
BTC to PI to be zero. The chainage of the BTC is 2371,87m.
Calculate data to orient the instrument at the BCC so that you would be able to set out the
portion from the BCC to the Crown Point from a zero instrument direction.
Calculate setting out data to set out the portion from the BCC to the Crown Point at 20m
continuous intervals.
Transition Curves 49
SSUR221
SURVEYING II
SUMMARY OF FORMULAE: TRANSITION CURVE
L
R
S
S
1.
2.
=
=
=
=
Lengths of Spirals
Radius of Circular Curve
Lengths to points on Spiral from BTC or ETC
Spiral Angle
S2
S =
rad
2 RL
x=S-
y=
L =
L
rad
2R
S5
S9
+
40 R 2 L2 3456R 4 L4
=
1718 , 87 L
deg
60R
when S = L, X = L -
S3
S7
S 11
+
6 RL 336 R 3 L3 42240R 5 L5
L3
L5
+
40 R 2 3456R 4
when S = L, Y =
L2
L4
L6
+
6 R 336R 3 42240R 5
S
3.
S2
572, 958 S 2
S =
=
rad =
deg
3
6 RL
60 RL
4.
Shift SH =
5.
Apex Distance AD = (R + SH) Tan
6.
Total Length
L2
L4
L6
−
+
24 R 2688R 3 506880R 5
Ι
L
L3
L5
+
+
2
2 240R 2 34560R 4
=
2L + R(  - 2L )
=
2L +
=
(R + SH) Sec
πR(Ι - 2φL )
180
when  and 2L are in rad
when  and 2L are in deg
Ι
-R
2
7.
Crown Distance
8.
Chord BCC – ECC = 2R sin (
9.
Chord lengths from BTC / ETC on the spiral = d = S −
 - 2L
)
2
Transition Curves 50
S5
S9
+
90 L2 R 2 22680L4 R 4
SSUR221
SURVEYING II
Chapter 4
DISTANCE MEASUREMENT
EMSH:
4.1
Chapter I (Pages 28 to 33)
Chapter IX (Page 136)
The Accuracy of EDM Instruments
Manufacturers of this equipment specify the accuracy of their instruments as follows:
1)
A standard error in the measurement e.g. ± 5mm. This figure gives the
supposed limits of a kind of sporadic unpredictable instrument constant which
is a function of the electronics and independent of the distance being
measured.
2)
A proportional error due to variations in frequency and the effects of
meteorological data e.g. ± 2 parts per million of the distance.
EXAMPLE:
An electronic distance measuring device is quoted as having an accuracy of ± (10mm
+ 5ppm). According to this accuracy, what would the standard error of a distance of
2500m be?
SOLUTION:
Standard error of distance =
(A) 2 + (B x D) 2
Where:
A = S.E. of instrument constant
B = S.E. of frequency which is dependent on distance.
D = Distance.
Sx =
(10) 2 + (5 x 2,5) 2
= ±16,01 mm.
4.2
Instrumental Errors
All EDM measurements are subject to the following errors.
Scale Error (or Frequency Drift)
This is due to the fact that the measuring frequencies do not correspond exactly with
the design value of the EDM instrument. The error is proportional to the distance
measured. Consequently, the effect is more noticeable on long lines and can sometimes be as high as 20-30 ppm for short-range instruments. This source of error is of
Distance Measurement 51
SSUR221
SURVEYING II
greater importance in the longer-range instruments using microwaves. Significant
errors in the short-range infra-red instruments are not common.
Zero Error (or Index Error) or Instrument & Prism Constant
This occurs if there are differences in the mechanical, electrical and optical centres of
the EDM instrument and reflectors, and includes the prism constant. It is therefore
due to a difference between the mechanically defined centres of instruments (and
reflector-when used) and their electrical (optical) centres. The error when present, and
not allowed for, produces an effect akin to a miscentering of the instrument by the
operator and is independent of range.
This error is of constant magnitude, and care must be taken to eliminate it. The value
of a zero error obtained from a calibration procedure usually applies to an instrument
and reflector and if the reflector is changed, the zero constant changes.
Cyclic Error (or Instrument Non-linearity)
This error varies in a periodic manner and such periodic errors are prevalent in
infra-red instruments. As the distance between the transmitters and reflector is
changes, the error will rise to a maximum, fall to a minimum, and so forth, over and
over again. To achieve maximum accuracy from a given EDM system, this cyclic
error must be evaluated.
This error can be investigated by measuring a series of known distances spread over
the measuring wavelength of the instrument. If a calibration curve of (observed measured) distances is plotted against distance and a periodic wave is obtained, the
EDM instrument has a cyclic error.
In most EDM instruments factory calibration is performed so that cyclic errors are
uniformly distributed. The total peak to peak error is then within specified limits.
From a practical standpoint, calibration for cyclic effects is rarely needed unless the
instrument is to be used for projects in which very high accuracies are required, e.g.
deformation surveys.
4.3
EDM Calibration
The determination of scale, zero and cyclic errors of EDM instruments is known as
calibration and can be carried out by a number of different methods which have a
varying degree of sophistication. For most site work, calibration is carried out using
baselines and techniques involving both unknown and known baseline lengths are
used. From the above we will only focus on establishing the Zero (Index) Error in
this course.
ZERO (INDEX) ERROR.
Distance Measurement 52
SSUR221
SURVEYING II
This may be detected and evaluated using a line measured directly (SO), and in parts
such as SI, S2, S3, etc. In such a case if the zero correction of the instrument is C
then:
C = S0 – (S1+S2+……..Si)
i-1
4.4
Geometric Corrections
Slope Correction
When the slope distance has been obtained from an EDM measurement, a slope
correction must be applied to it in order to obtain the equivalent horizontal distance. If
the EDM and theodolite are coaxial as in most total stations and the telescope is tilted
and pointed at the centre of the prism the correct slope distance is obtained no matter
how the prism is tilted and the horizontal distance can be calculated. For an integrated
total station, this calculation is carried out by the instrument's microprocessor and the
result displayed automatically.
Height above Datum Correction
When a survey is to be based on the National coordinate system, the line measured
must be reduced to its equivalent length at mean sea level (MSL). The correction is
negative unless a line below MSL is measured.
Scale Enlargement
A map projection provides a means of representing the curved surface of the Earth on
a plane surface so that coordinate grids can be defined and maps drawn. The relative
positions of points on the grid are altered slightly from their ground positions as a
result of using a projection to account for the curvature of the Earth. Therefore,
distances calculated from National coordinates will not, in some cases, agree with
their equivalent measured on site.
To convert measured distances to projection (or grid) distances the measured distance
is converted to its equivalent at MSL and a scale factor is used. The value of the scale
factor varies across the country.
Meteorological Corrections
Using EDM equipment, the measurement of distance is obtained by measuring the
time of propagation of electromagnetic waves through the atmosphere. Whilst the
velocity of these waves in a vacuum is known, its value will be reduced according to
the atmospheric conditions through which the waves travel at the time of
measurement.
The value of the correction is affected by the temperature, pressure and water vapour
content of the atmosphere as well as by the wavelength of the transmitted
electromagnetic waves. It follows from this that measurements of these atmospheric
conditions are required at the time of measurement.
Distance Measurement 53
SSUR221
SURVEYING II
EDM equipment is standardized under certain conditions of temperature and pressure.
For instance, some makes of equipment are standardized at 20°C and 1013.25 mbar of
pressure, whilst others are standardized at 12°C and 1013.25 mbar. Under these
respective atmospheric conditions, the measured distances would not require a
velocity correction. It follows that even on low-order surveys the measurement of
temperature and pressure is important.
EXAMPLE
Reduce the given distances which were measured with an EDM. Corrections for
meteorological effects, height above mean sea level, scale enlargement, slope and the
instrument constant (Zero or Index Error) must be applied.
List of Formulae:
Meteorological correction: +0,020m/1000m
Slope: C = S(1 – Cos α )
MSL: C = SH/R
Scale enlargement: C = SY²/2R²
Instrument Constant: See next page
The mean height of the survey: 2200m
Assume the radius of the earth to be 6367km
The mean Y-value of area: -60 000,00m
Distance Measurement 54
SSUR221
SURVEYING II
AD = s = 271,511
AC = t = 150,404
AB = u = 50,477
CB = v = 100,065
CD = w = 121,252
Calculation of inst constant:
E = ½ (u+v+w-s) = +0,142
E = u+v-t
= +0,138
E = t+w-s
= +0,145
MEAN
= +0,142
CORRECTION = -0,142
SOLUTION
FIELD DATA
CORRECTIONS
RED
DIST
STA
MEAS.
DIST.
VERT.
ANGLE
MET
SLOPE
MSL
SCALE
INST
CONST
A-B
264,49
-2.11
+0,005
-0,192
-0,091
+0,012
-0,142
264,082
B-C
201,89
-3.01
+0,004
-0,280
-0,070
+0,009
-0,142
201,411
C-D
5,07
+11.17
0
-0,098
-0,002
0
-0,142
4,828
D-E
302,83
+1.56
+0,006
-0,172
-0,105
+0,013
-0,142
302,430
Distance Measurement 55
SSUR221
SURVEYING II
EXERCISES
1.
Reduce the measured distances so that they can be used in a traverse calculation. You
can complete the question on the next page.
Data: Distances measured to obtain zero (index) error:
AB = 170,515m
BC = 140,370m
CD = 215,658m
AC = 310,852m
AD = 526,475m
•
Mean height of the traverse: 2200m. Assume R = 6367km.
•
Mean Y-value of area: -60 000,00m
•
Temp during measurement was 33°C. The pressure was 780mb throughout. Use the
attached nomogram for meteorological corrections.
Nomogram for corrections in PPM or mm/1000m
Distance Measurement 56
SSUR221
SURVEYING II
Calculation of zero (index) error:
Sketch:
Calculation:
Calculation of reduced distances:
Distance Measurement 57
SSUR221
SURVEYING II
Formulae:
Slope: C = S(1 - cosα )
MSL: C = SH/R
Scale enlargement: C = SY²/2R²
Measured
Distance
2 124,66m
Vert
Angle
+3°15'
3 077,42m
-1°30'
1 891,43m
+0°45'
2.
Slope Corr MSL Corr Scale Corr
Met
Corr
Zero Corr
Reduced
Distance
Reduce the given distances, which were measured with an EDM. Corrections for
meteorological effects, height above mean sea level, scale enlargement, slope and the
instrument constant (Zero or Index Error) must be applied.
List of Formulae:
Meteorological correction: +0,002m/100m
Slope: C = S(1 - cos )
MSL: C = SH/R
Scale enlargement: C = SY²/2R²
Instrument Constant (zero error): +0,142m
The mean height of the traverse: 2500m
Assume the radius of the earth to be 6367km
The mean Y-value of area: -65 000,00m
Calculation of distances:
FIELD DATA
CORRECTIONS
Distance Measurement 58
REDUCED
SSUR221
SURVEYING II
DISTANCE
STATION
MEAS.
DIST.
VERT.
ANGLE
A-B
246,23
-2º 11´
B-C
241,33
-3º 41´
C-D
5,11
+11º 27´
D-E
321,86
+1º 36´
3.
MET
SLOPE
MSL
SCALE
INST
CONST
Reduce the measured distances so that they can be used in a traverse calculation.
Calculation of reduced distances: Formulae:
▪
▪
▪
▪
▪
▪
▪
▪
Meteorological correction is 38mm per 1000m
Slope: C = S(1 - cos )
Instrument Constant (zero error): +0,034m
MSL: C = SH/R
The mean height of the traverse: 1200m
Scale enlargement: C = SY²/2R²
Assume the radius of the earth to be 6367km
The mean Y-value of area: -70 000,00m
Measured
Distance
Vert
Angle
2 122,66m
+2°15'
3 075,42m
-2°30'
1 896,43m
+1°45'
Slope Corr MSL Corr Scale Corr
SOLUTIONS
EXERCISE 1
Distance Measurement 59
Met
Corr
Zero Corr
Reduced
Distance
SSUR221
SURVEYING II
Reduce the measured distances
E=
E=
E = s+w-t
½ (u+v+w-t) = +0,034
u+v-s
= +0,033
= +0,035
MEAN
= +0,034
Calculation:
Correction = - 0,034m
Met Corr from nomogram is 80mm per 1000m
Calculation of reduced distances:
Measured
Distance
Vert
Angle
Slope
Corr
MSL
Corr
Scale
Corr
Met
Corr
Zero
Corr
Reduced
Distance
2 124,66m
+3°15'
-3,417
-0,734
+0,094
+0,170
-0,034
2120,739m
3 077,42m
-1°30'
-1,055
-1,063
+0,137
+0,246
-0,034
3075,651m
1 891,43m
+0°45'
-0,162
-0,654
+0,084
+0,151
-0,034
1890,815m
EXERCISE 2
FIELD DATA
CORRECTIONS
REDUCED
DISTANCE
STATION
MEAS.
DIST.
VERT.
ANGLE
MET
SLOPE
MSL
SCALE
INST
CNST
A-B
246,23
-2º 11´
+0,005
-0,179
-0,097
+0,013
-0,142
245,830
B-C
241,33
-3º 41´
+0,005
-0,499
-0,095
+0,013
-0,142
240,612
C-D
5,11
+11º 27´
0
-0,102
-0,002
0
-0,142
4,864
D-E
321,86
+1º 36´
+0,006
-0,125
-0,126
+0,017
-0,142
321,490
EXERCISE 3
Distance Measurement 60
SSUR221
SURVEYING II
Measured
Distance
Vert
Angle
Slope
Corr
MSL
Corr
Scale
Corr
Met
Corr
Zero
Corr
Reduced
Distance
2 122,66m
+2°15'
-1,637
-0,400
+0,128
+0,081
-0,034
2120,798m
3 075,42m
-2°30'
-2,927
-0,580
+0,186
+0,117
-0,034
3072,182m
1 896,43m
+1°45'
-0,885
-0,357
+0,115
+0,072
-0,034
1895,341m
.
Distance Measurement 61
[Date]
Chapter 5
TRAVERSING USING A TOTAL STATION
EMSH:
5.1
Chapter VIII (Pages 123 to 130)
Booking observations in the field
It is assumed that the procedure for setting up the total station is covered during the practical
sessions. In these notes special mention will be made of a few precautions to be taken when
using a total station.
•
Checking the optical plummet
The method will depend upon whether the plummet optics are situated in the alidade or in the
tribrach.
If the former, the field method is to set-up over hard flat ground and level carefully. Fix a
piece of paper on the ground below the instrument and make a mark on it to coincide with the
optical axis of the plummet. Rotate the alidade through 180° about the vertical axis. The
plummet axis will also be rotated and if it is vertical, will still pass through the mark on the
paper. If it is out of adjustment however, it will give a new intersection with the paper which
should be marked. A point mid-way between the two marks is the point through which the
plummet axis should pass, and this should also be marked.
Usually the plummet is in the tribrach, the previous method cannot be used because the
optical axis of the plummet cannot be rotated through 180° without disturbing the levelling.
One way of carrying out the adjustment in the field is to suspend a plumb bob from the total
station and then to set the plummet against the plumbed mark. However this method is not
really suitable since the plummet should be more accurate than the plumb bob. An alternative
method is to set-up the total station above a dish of mercury, thereby making use of the
principle of auto-collimation. The image of the centre of the plummet objective should
coincide with the intersection of the cross-hairs on the plummet reticule. If it does not, then
the plummet is adjusted until coincidence is obtained. Careful levelling is important.
Another method which does not require a dish of mercury is to clamp the telescope roughly
horizontal and then to place the total station on a table so that it rests on the vertical circle
casing, the other standard and the telescope objective. In this position the total station should
be supported so that its vertical axis is more or less horizontal and so that the tribrach can be
rotated about the vertical axis relative to the alidade. The test and adjustment can then be
carried out as described earlier, but this time the intersections of the optical axis with a card
attached to the wall are marked and the plummet adjusted to the central position. The card
should be approximately normal to the axis of the plummet.
A very useful method to be used in the field is to mark the position of the tribrach on the
tripod head by using a soft-point erasable marker pen. Unclamp the tribrach and turn through
120 degrees, so that it coincides with the position marks. Level the tribrach and check if the
optical plummet still bisects the peg. If not, the optical plummet is out of adjustment.
Traversing 62
[Date]
•
Other precautions
Most modern total stations are equipped with on-board software which will apply corrections
to measured distances. It is important for the surveyor to know exactly what the status of the
data is that is booked. Often values for atmospheric pressure and temperature are stored and
the total station will automatically apply corrections for meteorological effects. Similarly the
Instrument and Prism Constant, height above datum, scale enlargement and slope angles can
automatically be corrected for and a reduced distance is displayed.
If the surveyor is unaware that these corrections have already been applied by the instrument,
the danger exists that they could be applied again after the distances have been measured.
5.1.1 Orientation
Once the Total Station has been set up over the starting point of the traverse – a point
with known coordinates – calculate joins to at least two other visible known points for
orientation. Select the longest ray, which in addition is well defined, as your reference
object (RO), and bisect the RO. Then follow the prescribed procedure for the make/
model of total station you are using to set the join direction into the instrument.
Check your orientation by bisecting the other known point which is visible from your
instrument position. Compare this direction with the join direction, which should not
differ by more than a few seconds.
5.1.2 The observation procedure (manual booking)
Revision – Chapter 1
5.1.3 Reduction of the field book
Revision – Chapter 1
5.2
The traverse direction sheet
Revision – Chapter 1
5.3
The Traverse Calculation
Revision – Chapter 1
Traversing 63
SSUR221
SURVEYING II
Chapter 6
THE DETERMINATION OF HEIGHTS
1. Trigonometric Levelling
B
h

Horizontal through centre of instrument
Horizontal through ground point A
HI
C&R
A
Surface parallel to earth’s surface
δh = ± h’ +HI +C&R –HS
In the sketch above, HS (height of signal) is zero. Vertical observations to trig beacons are
usually done by observing the top of the concrete pillar, which is the known height.
•
•
•
•
•
h’ will always have the same sign as α, and the sign must be written down ay all times
δh will not necessarily have the same sign as α. Where h’ is small in relation to the
other terms, it may have the opposite sign
HI = Height of Instrument above point A
HS = Height of the line of sight at the signal
C&R = combined effect of curvature and refraction
1.1. Curvature correction
The curvature correction is introduced because the surface of the earth is not a
horizontal plane. It is the departure of the tangent to the earth’s surface at the point of
observation from the imaginary curve representing the earth’s surface. This departure
S2
is equal to
plus another term so small that it can be left out for most practical
2R
purposes.
R = the radius of the earth at the mean latitude of the two end points.
Determination of Heights 64
SSUR221
SURVEYING II
S = the length of the line, as found from coordinates. For precise work, S must be
H1 + H2
multiplied by 1 +
to give an elevated distance related to the mean sea level
2R
of the given line. This is indicated by S’.
OR
 H 
S’ = S 1 + m  where H m is the mean height above datum of the area of the survey.
R 

1.2. Refraction
As a result of refraction the line of sight is not straight but is concave to the earth’s
surface. It is a very uncertain quantity because the value changes as the atmospheric
conditions change, causing trig levelling to be rather uncertain. Since the error due to
refraction can only be determined at any given time by actual experiments, a mean
value has been adopted for South Africa. This has been combined with curvature and
the general formula is changed to:
C&R =
1− k
2
x (S' )
2R
where k is the mean refraction constant with an approximate value of 0, 13.
Experience has shown that refraction is most constant during the middle of the day
and very variable early morning and late afternoon.
The full calculation for δh is then as follows:
δh = S’ tan α +
1− k
1− k
2
2
x (S' ) tan 2 α
x (S' ) + HI – HS +
2R
2R
The last term is only for very steep sights and will not be used in this module.
So δh = S’ tan α +
1− k
2
x (S' ) + HI – HS
2R
In this module we will only consider single point fixes where observations from or to
one point only. We will not be doing trig levelling networks.
PTO for example
EXAMPLE – Trigonometrical Levelling
Calculate the elevation of Q. Take the correction for refraction as one seventh of the
correction for curvature. Take R = 6367km.
Determination of Heights 65
SSUR221
@Q
SURVEYING II
HI = 1,67m
Obs to:
CL
CR
Description
Join distances
R
88.57.29
271.02.50
Top flag 2, 03m
2142, 16m
S
92.02.57
267.56.39
Top concrete
1071, 08m
T
87.12.40
272.47.40
Top flag 1, 67m
1606, 62m
R
1788,29
Peg at ground level
S
1712,54
Trig Beacon, top of concrete
T
1827,83
Peg at ground level
Elevations:
SOLUTION
 Hm 
1 +
 = 1,00028
R 

1
(given)
7
STATION
R
S
T
CL 
+01.02.31
-02.02.57
+02.47.20
CR 
+01.02.50
-02.03.21
+02.47.40
Mean 
+01.02.40
-02.03.09
+02.47.30
S
2142,16
1071,08
1606,62
S'
2142,76
1071,39
1607,07
S' tan 
+39,06
-38,40
+78,37
C+R
+0,31
+0,08
+0,17
H.I.
+1,67
+1,67
+1,67
H.S.
-2,03
0
-1,67
Σ= h
+39,02
-36,65
+78,54
Elevation
1788,29
1712,54
1827,83
Q
1749,27
1749,19
1749,29
Weight 1/S²
22
87
39
Weighted mean
height of Q
EXERCISES
1.
k=
1749,23
From the following data calculate the weighted mean elevation of the ground peg P,
giving your answer to the nearest 0, 01 of a metre.
Determination of Heights 66
SSUR221
SURVEYING II
Vertical Angle Observations
@P
(H.I. = 1,59m)
C.L.
89.08.00
88.54.45
96.46.00
A
B
C
Join Dist. (m)
C.R.
270.51.46
271.05.00
263.13.40
Top of Signal 0,92m high
Top of concrete pillar
Top of concrete pillar
Elevations (m)
P – A 7579,88m
P – B 6712,00m
P – C 343,76m
A
B
C
828,44m
841,24m
670,25m
SOLUTION: Assume for this example: 1 + Hm/R = 1,00122503
Radius = 6367 km
STATION
C & R = 0,435S²
R
A
B
C
CL 
0.52.00
1.05.15
-6.46.00
CR 
0.51.46
1.05.00
-6.46.20
Mean 
0.51.53
1.05.08
-6.46.10
S
7579.88
6712.00
343.76
S'
7589.166
6720.222
S' tan α
C+R
H.I.
H.S.
Σ= h
Elevation
P
Weight 1/S²
Weighted mean
height of P
2.
709.06
In order to do a power-line profile it was necessary to fix the initial point by trig
heighting using single height differences. Calculate the weighted mean elevation of
the ground point GQUMA giving your answer to the nearest 0.01 of a metre.
Determination of Heights 67
SSUR221
SURVEYING II
Vertical Angle Observations
@ GQUMA (H.I. = 1,62m)
C.L.
88.39.10
93.05.28
89.10.05
Flagstaff
St Elmos
Lusikisiki
C.R.
271.21.05
266.54.21
270.50.12
Join Dist. (m)
Gquma - Flagstaff
Gquma - St Elmos
Gquma - Lusikisiki
Top of Signal 1,20m high
Top of Church Spire
Top of Pillar
Elevations (m)
5452,27
1888,85
6005,42
Flagstaff
St Elmos
Lusikisiki
1010,54
779,51
971,24
SOLUTION
Assume for this example: 1 + Hm/R = 1,000144
Radius = 6373 km
STATION
C & R = 0,435S²
R
Flagstaff
St Elmos
Lusikisiki
CL 
CR 
Mean 
S
S'
S' tan α
C+R
H.I.
H.S.
Σ= h
Elevation
Gcuma
Weight 1/S²
Weighted mean
height of Gcuma
SOLUTIONS OF EXERCISES – TRIG. LEVELLING
1.
From the following data calculate the weighted mean elevation of the ground peg P,
giving your answer to the nearest 0, 01 of a metre.
Determination of Heights 68
SSUR221
SURVEYING II
Vertical Angle Observations
@P
(H.I. = 1,59m)
C.L.
89.08.00
88.54.45
96.46.00
A
B
C
Join Dist. (m)
C.R.
270.51.46
271.05.00
263.13.40
Top of Signal 0,92m high
Top of concrete pillar
Top of concrete pillar
Elevations (m)
P – A 7579,88m
P – B 6712,00m
P – C 343,76m
A
B
C
828,44m
841,24m
670,25m
SOLUTION: Assume for this example: 1 + Hm/R = 1,00122503
C & R = 0,435(S’)²
R
Radius = 6367 km
STATION
A
B
C
CL 
+00.52.00
+01.05.15
+06.46.00
CR 
+00.51.46
+01.05.00
-06.46.20
Mean 
+00.51.53
+01.05.08
-06.46.10
S
7579,88
6712,00
343,76
S'
7589,17
6720,22
344,18
S' tan α
+114,55
+127,34
-40,85
C+R
+3,93
+3,09
+0,01
H.I.
+1,59
+1,59
+1,59
H.S.
-0,92
0
0
Σ= h
+119,15
+132,02
-39,25
Elevation
828,44
841,24
670,25
P
709,29
709,22
709,50
Weight 1/(S')²
17
22
8442
Weighted mean
height of P
2.
709,50
In order to do a power-line profile it was necessary to fix the initial point by trig
heighting using single height differences. Calculate the weighted mean elevation of
the ground point GQUMA giving your answer to the nearest 0.01 of a metre.
Vertical Angle Observations
Determination of Heights 69
SSUR221
SURVEYING II
@ GQUMA (H.I. = 1,62m)
C.L.
88.39.10
93.05.28
89.10.05
Flagstaff
St Elmos
Lusikisiki
C.R.
271.21.05
266.54.21
270.50.12
Join Dist. (m)
Gquma - Flagstaff
Gquma - St Elmos
Gquma - Lusikisiki
Top of Signal 1,20m high
Top of Church Spire
Top of Pillar
Elevations (m)
5452,27
1888,85
6005,42
Flagstaff
St Elmos
Lusikisiki
1010,54
779,51
971,24
SOLUTION
Assume for this example: 1 + Hm/R = 1,000144
Radius = 6373 km
C & R = 0,435(S’)²
R
STATION
Flagstaff
St Elmos
Lusikisiki
CL 
+01.20.50
-03.05.28
+00.49.55
CR 
+01.21.05
-03.05.39
+00.50.12
Mean 
+01.20.58
-03.05.34
+00.50.04
S
5452,27
1888,85
6005,42
S'
5453,055
1889,122
6006,285
S' tan α
+128,456
-102,072
+87,479
C+R
+2,030
+0,244
+2,462
H.I.
+1,62
+1,62
+1,62
H.S.
-1,20
0
0
Σ= h
+130,906
-100,207
+91,563
Elevation
1010,54
779,51
971,24
Gcuma
879,634
879,717
879,677
Weight 1/(S')²
34
280
28
Weighted mean
height of Gcuma
2.
879,71
Height traversing
It is often necessary when fixing points in the horizontal to also calculate vertically the
heights of the control points. The observations for this must be done simultaneously when
doing the traverse. Here vertical angles and distances are used to determine δh for each leg
of the traverse. With the advent of the EDM and more recently the electronic theodolite
Determination of Heights 70
SSUR221
SURVEYING II
(total station) this has become a popular method of obtaining the heights of working points.
This method is fairly reliable, and as long as a thorough field procedure is adhered to, then an
accuracy of 1cm to 2cm can be obtained.
Obviously if greater accuracy (mm) is required spirit levelling would be the reliable and
necessary route to follow. The question here is when is height traversing used?
*
Most common when traversing for a detail survey because it is a very
economical means of obtaining heights for working points.
*
It also saves time when bench marks are not readily available -as spirit
levelling can be a time consuming exercise.
*
It is also used for application plans - subdivisions
*
It can also be used for vertical control for profiling, especially for electrical
power lines.
*
It can be used quite satisfactorily for construction purposes e.g. the setting out
of batter/ profile boards in controlling earthworks, digging trenches, etc.
However if mass concrete work is involved (bridge decks) or critical grades
are to be maintained (outfall sewers), it is necessary to work more accurately,
and spirit leveling will be the route to follow.
Obviously observation procedures differ depending on whether one is using mechanical or
electronic theodolites. With the former one needs to measure the HI, the slope distance, the
vertical angle and the HS, where as with electronic theodolites one has the option to do the
same or to measure HI, the height diff )h and the HS. What follows is a description of the
first method while the second method follows the same principles but does not involve
having to calculate )h.
Field Procedure and Method of Calculation
Determination of Heights 71
SSUR221
SURVEYING II
REVISION: SURVEYING I: VERTICAL INDEX ERROR
See above section for theory on the Vertical index Error. This is of great significance when
doing a heighted traverse.
This error is derived as follows:
Subtract the sum of CL and CR from 360º. The difference will represent twice the index
error.
Example 1 – observing upwards
Example 2– observing upwards
CL
85.26.20
CL
86.51.17
CR
274.37.20
CR
272.57.33
Σ = 360.03.40
Σ = 359.48.50
I.E. = (360 – 360.03.40) ÷ 2
I.E. = (360 – 359.48.50) ÷ 2
 I.E. = -00.01.50
 I.E. = +00.05.35
Corrected readings are:
Corrected readings are:
CL: 85.26.20
CR: 274.37.20
CL: 86.51.17
CR: 272.57.33
-00.01.50
-00.01.50
+00.05.35
+00.05.35
85.24.30
274.35.30
86.56.52
273.03.08
Σ CL & CR = 360.00.00
Σ CL & CR = 360.00.00
As was stated in Chapter 3, the index error can be completely eliminated by taking the mean
of the CL and CR vertical angles.
Example 1
Example 2
CL = 85.26.20
α = +04.33.40
CL = 86.51.17
α = +03.08.43
CR = 274.37.20
α = +04.37.20
CR = 272.57.33
α = +02.57.33
Mean α
+04.35.30
Mean α
+03.03.08
Check:
Check:
CL: 90 - 04.35.30
85.24.30
CL: 90 - 03.03.08
86.56.52
CR: 270 + 04.35.30
274.35.30
CR: 270 + 03.03.08
273.03.08
Σ CL & CR = 360.00.00
Σ CL & CR = 360.00.00
Determination of Heights 72
SSUR221
SURVEYING II
Example 3 – observing downwards
Example 4 – observing downwards
CL
93.16.05
CL
92.20.37
CR
266.42.40
CR
267.38.01
Σ = 359.58.45
Σ = 359.58.38
I.E. = (360 – 359.58.45) ÷ 2
I.E. = (360 – 359.58.38) ÷ 2
 I.E. = +00.00.37,5
 I.E. = +00.00.41
Corrected readings are:
Corrected readings are:
CL: 93.16.05
CR: 266.42.40
CL: 92.20.37
CR: 267.38.01
+00.00.37,5
+00.00.37,5
+00.00.41
+00.00.41
93.16.42,5
266.43.17,5
92.21.18
267.38.42
Σ CL & CR = 360.00.00
Σ CL & CR = 360.00.00
Using vertical angles:
Example 3
Example 4
CL = 93.16.05
α = -03.16.05
CL = 92.20.37
α = -02.20.37
CR = 266.42.40
α = -03.17.20
CR = 267.38.01
α = -02.21.59
Mean α
-3.16.42,5
Mean α
-2.21.18
Check:
Check:
CL: 90 + 3.16.42,5
93.16.42,5
CL: 90 + 2.21.18
92.21.18
CR: 270 -3.16.42,5
266.43.17,5
CR: 270 - 2.21.18
267.38.42
Σ CL & CR = 360.00.00
Σ CL & CR = 360.00.00
Calculating height differences
* The vertical reading must be read to the nearest second and both CL and CR must be
read.
*
A height difference must be calculated both forwardly and backwardly and therefore
the HS and the vertical reading to the backward point must also be measured.
*
For distances longer than 100m the curvature and refraction correction must be applied.
C&R corrn = 0,435xS²
R
Refer to the sketch above and calculate the mean height diff. as follows:Determination of Heights 73
SSUR221
SURVEYING II
Ht. diffAB = HI + S cos Z + C&R - HS
Example = 0,233 +200, 00 x cos9515' + 0,003 – 1,585 = -19,649
Or if using vertical angles:
= 0,233 + 200, 00 x sin (-5º 15’) + 0,003 – 1,585 = -19,649
Ht. diffBA = HI + S' cos Z + C&R – HS
Example = 1,685 +199, 96 x cos8451'+ 0,003 – 0, 00
= +19,637
Or if using vertical angles:
= 1,685 + 199, 96 x sin (+5º 09’) +0,003 – 0, 00 = +19,637
Mean height difference AB = -19,643
Continuing with B – C ……..
Ht. diffBC = HI + S cos Z + C&R – HS
Example = 1,685 +130, 00 x cos9310' + 0,001 -1,495
= - 6, 990
Or if using vertical angles:
= 1,685 + 130, 00 x sin (-3º 10’) + 0,001 -1,495 = -6,990
Ht. diffCB = HI + S cos Z + C&R - HS
Or if using vertical angles:
Ht. diffCB = HI + S sin (±α) + C&R - HS
ETC.
•
•
•
•
•
•
•
S is the slope distance.
S' is the horizontal distance obtained from BA which is used to reduce the back height
diffBA using S'cosZ.
Z is the vertical angle READING (reduced to CL after applying index error).
Α is the mean vertical angle: + FOR UPWARD OBSERVATIONS AND - FOR
DOWNWARD OBSERVATIONS.
C&R is the curvature and refraction correction.
HI is the height of instrument to the nearest millimetre.
HS is the height of signal (target) also to millimetres.
EXAMPLE
Determination of Heights 74
SSUR221
SURVEYING II
See the tabulated results of a height traverse below. Use this data to calculate the final heights of
the intermediate traverse stations.
STATION
HORI DIST
HT. DIFF
RED. HT.
ADJUST
FINAL HT.
T6
34.720
65,610
-3,012
111,982
+0,172
190,864
+5,056
146,442
+11,242
162,261
+13.465
244,490
+13,212
212,357
+23,452
FR1
FR2
INTER
FR3
FR4
FR5
98.340
SPORT
Σ= 1134,006
The misclosure is adjusted proportional to the distance, i.e. the longer distance gets a bigger
correction.
SOLUTION: HEIGHT TRAVERSE AND ADJUSTMENT
Determination of Heights 75
SSUR221
SURVEYING II
STATION
HORI DIST
HT. DIFF
RED. HT.
ADJUST
T6
FINAL HT.
34,720
65,610
-3,012
FR1
111,982
SPORT
31,885
36.936
+0,011
36,947
48.178
+0,015
48,193
61.643
+0,020
61,663
74.855
+0,027
74,882
98.307
+0,033
98,340
+13,212
FR5
212,357
+0,005
+13.465
FR4
244,490
31.880
+11,242
FR3
162,261
31,710
+5,056
INTER
146,442
+0,002
+0,172
FR2
190,864
31.708
+23,452
Σ= 1134,006
Adjustment
T6 – FR1 = 65,610/1134, 006 x 0, 033 = 0,002
FR1 – FR2 = [65, 61+111,982]/1134, 006 x 0, 033 = 0,005
ETC
Try the rest here:
Height Traverse & Adjustment – further worked example.
Determination of Heights 76
SSUR221
STATION
SURVEYING II
HORI DIST
HT DIFF.
HEIGHT
ADJUST.
Δ ALOES
FINAL HEIGHT
371,500
431,21
-52,312
T1
652,42
319,208
282,692
+0.050
282,742
253,082
+0.110
253,192
298,673
+0.127
298.800
-29,610
T3
371,98
+0.020
-36,496
T2
1286,35
319,188
+45,591
Δ EUPH
Σ= 2741,96
Again, try the adjustment calculations here:
Determination of Heights 77
SSUR221
SURVEYING II
Chapter 8
CONSTRUCTION SURVEYING: SETTING OUT OF BUILDINGS
EMSH:
Chapter XI (Pages 155 to 156)
Construction Surveying consists of all forms of setting out as well as the survey of existing
buildings, and encompasses a number of the other facets of survey i.e. linear and angular
measurements (traversing), tacheometry and levelling. It must be mentioned that it can be
very intricate e.g. jacking of a bridge deck in suspension, welding of super-structures for the
Moss-Gas Project, tunnelling, setting up monitors for robot controlled motor assembly lines
etc., and varied e.g. laying of underground sewers, monitoring of deformations for dams,
seismic activity, measuring of bulk materials (earthworks) etc. Some of these are skills which
often demand specialists.
Obviously it is well near impossible to go into detail on all aspects of the above mentioned
surveys; however for the purposes of this course we will cover some of the more day to day
survey activities common to construction survey. Due to the varying nature of construction
surveys it is proposed to cover them under specific headings.
Mention must also be made, that most often setting out is done by simple polar and with the
advent of total stations, and both horizontal and vertical positions can be controlled three
dimensionally. More recently on-board computers allow coordinated data to be stored, so
that setting out has become more simplified. Data Logging Programmes are specifically
designed for Construction Surveys and have a number of very clever features.
Construction of buildings
Setting out from property beacons and building lines
Most building sites are governed by the extremities of the property on which they are to be
built. It is strongly recommended that before any building commences the property beacons
of the site in question be exposed and checked by a professional surveyor, especially if you
are going to build right up to the boundary line, or if the building lines as stipulated by the
local authority have to be adhered to. If the beacons for a particular area have been recently
placed and are clearly visible (next to a marker), or the owner of the property has indicated
their positions, it is still necessary to conduct the following checks:1)
Obtain a copy of the survey diagram or general plan; this will give the site dimensions
and the beacon descriptions.
2)
Measure between the beacons to see whether it corresponds to those on the diagram/
general plan. If you have the use of a total station and you are able to set up over one
of the beacons all the better. The measured radial polars should then be carefully
compared with the join data for consistency.
Also check to see whether the physical features of a beacon in the field correspond to
the beacon description indicated on the diagram/ general plan.
3)
Setting Out 103
SSUR221
4)
SURVEYING II
If there is any discrepancy, or any chance of serious repercussions you are advised to
call in a Professional Land Surveyor.
Always check that you use an approved Diagram or General Plan. Never, repeat, never take
the site dimensions off a building plan. Once the beacons have been established (checked),
the building lines may be set out either by parallel lines (if the property is rectangular), or if
not, by co-ordinates.
Setting out of building lines
The sketch above indicates typical building lines for a residential site.
Alternatively it is often necessary to set out the position of the house as positioned by the
architect or plan draughts person. They may very well have skewed the building at an angle
to accommodate for the slope of the land or make the front of the house north facing
(orientation) or to accommodate for a particular view (aspect). It is normally only possible
to do this when the property is of a fair size (>1500m²), or slightly smaller (1000m²) if it is a
level site. An example can best illustrate this (see sketch below).
Setting Out 104
SSUR221
SURVEYING II
Setting out can be done by in number of ways; either by placing a peg at the required distance
along AB and then turning of the angle (say 110º); or measuring the distance along the
opposite boundary DC so as to create a base line; or if a grid is provided, scaling off the coordinates of the corners of the building and then placing them by polars from, say, A.
At all times it is necessary to check the setting out of corner pegs, by check taping between
pegs, also diagonals. However if the site needs earthworks it is customary to do these first. It
is standard practice to make the site platform bigger than the extent of the building (+/-1,5m
all round to allow for the erection of scaffolding etc.) In fact if the site allows an even bigger
platform it should be constructed, to allow for more leisure space or later additions.
Setting Out 105
SSUR221
SURVEYING II
Foundations - corner profiles and widths, levels and stepping
The corner pegs are first set out from the corner beacons using a total station. The corner pegs
are then checked (diagonals) with a tape. Offset pegs are now placed, generally 1m or 2m
away at right angles to the building, two per corner. (See sketch below).
EXAMPLE 1
See Survey Diagram of Portion 2 of Erf 213 Forest Hills.
Calculate setting out data for a rectangular building 30m x 15 m which is to be erected on this
site. The 30m side of the building will be parallel to boundary DE, and the 15 side will be
parallel to boundary EF. The building line for both boundaries is 5 metres. All four of the
building corners must be marked with pegs at 2 metre offsets. Calculate coordinates for all
eight offset pegs. HINT: Start by calculating the coordinates of the corner of the building near
point E, where the boundary forms a 90º angle.
Setting Out 106
SSUR221
SURVEYING II
Setting Out 107
SSUR221
SURVEYING II
SOLUTION
Dir DE = 200.20.30 (from diagram)
Dir EF = 290.20.30 (from diagram)
Coordinates from Diagram:
D
+15 401,02 +2 851,52 E
+15 370,41 +2 768,94
F
+15 340,21 +2 780,41
Polars:
E – X1:
+15 365,722 +2770,678
X1 – C1:
+15 367,460 +2775,366
E – X2:
+15 372,148 +2773,628
X2 – C1:
+15 367,460 +2775,366
C1A: C1B:
+15 366,765 +2773,491
+15 369,335 +2774,671
E
C1A
X1
F
5m
X2
5m
C1
D
Setting Out 108
C2A
SSUR221
SURVEYING II
C2:
C2A:
C2B:
+15 353,395 +2780,581
+15 352,700 +2778,705
+15 351,520 +2781,276
C3:
C3A:
C3B:
+15 363,824 +2808,710
+15 364,519 +2810,585
+15 361.949 +2809,405
C4:
C4A
C4B
:C1:
+15377,888 +2803,495
+15 378,584 +2805,371
+15 379,764 +2802,800
+15 367,460 +2775,366
C2B
C1B
C1
C2
C4
C3
C3B
C4B
EXAMPLE 2
Below is an extract of the Survey Diagram of
Portion 7 (of 2) of Erf 65 Amanzimtoti.
C4A
C3 A
Calculate setting out data for a rectangular building 30m x 15 m which is to be erected on this
site. The 30m side of the building will be parallel to boundary EA, and the 15 side will be
parallel to boundary DE. The building line for both boundaries is 5 metres. All four of the
building corners must be marked with pegs at 2 metre offsets. Calculate coordinates for the
four corners of the building. YOU DO NOT HAVE TO CALCULATE COORDINATES
FOR THE OFFSET PEGS.
HINT: Start by calculating the coordinates of the corner of the building near point E, where
the boundary forms a 90º angle.
Setting Out 109
SSUR221
SURVEYING II
SOLUTION
E
X1
A
5m
X2
5m
C1
D
Polars:
Direction
Distance Name
Y
X
————————————————————————————————————————————————————————
Setting Out 110
SSUR221
SURVEYING II
5,000
E
2382,330
6207,270
X1
2378,034
6204,711
=================================
329:13:10
5,000
E
2382,330
6207,270
X2
2379,771
6211,566
=================================
329:13:10
5,000
X1
2378,034
6204,711
C1
2375,476
6209,007
=================================
239:13:10
239:13:10
X2
2379,771
6211,566
5,000 C1
2375,476
6209,007
Previous C1
2375,476
6209,007
Adopted
C1
2375,476
6209,007
=================================
239:13:10
30,000
C1
2375,476
6209,007
C2
2349,702
6193,654
=================================
329:13:10
15,000
C2
2349,702
6193,654
C3
2342,025
6206,541
=================================
59:13:10
30,000
C3
2342,025
6206,541
C4
2367,799
6221,894
=================================
149:13:10
15,000
Previous
Adopted
C4
2367,799
6221,894
C1
2375,476
6209,007
C1
2375,476
6209,007
C1
2375,476
6209,007
=================================
Setting Out 111
SSUR221
SURVEYING II
Chapter 9
VERTICAL CURVES
EMSH:
Chapter XI (Pages 181 to 186)
SUMMARY OF FORMULAE: SYMMETRICAL VERTICAL CURVE
1.
G = g 2 − g1
2.
e=
LG
800
3.
y=
(x 1 )2 e
( 1 )2
NOTE:
1 =  2
For max value of h
x=
g1 ( 1 )
200 e
2
or x =
g 2 ( 2 )
200 e
2
 1 +  2 = L = curve length BVC - EVC
EXAMPLE 1
A symmetrical parabolic summit (crest) vertical curve is to link two gradients of a proposed
new road. The height of the point of intersection of the straights is 857,93m. The tangent
lengths are each 150m long at gradients of +5% and -3,0% respectively. Calculate the
reduced levels of the curve at 20 m intervals, on continuous chainages of 20m. The chainage
of the VPI is 350m. Also calculate the position and height of the highest point on the curve.
G = g 2 − g 1 = (-3, 0) -5 = -8, 0 % (Grade Angle)
Vertical Curves 111
SSUR221
e=
SURVEYING II
300x8
LG
=
= 3,000m (Numerical value)
800
800
Calculation of Staking Data
In calculating vertical curves we require the reduced levels of a series of pegs usually placed
at 20m intervals along the centre line of the road.
If a tangent with a gradient of + g 1 and length  1 is divided into a number of equal parts, each
x metres long, the levels of the successive points along the tangent will be, starting at zero
height for the first peg:
g1. x
100
NOTE:
,
g1. 2 x
100
,
g1. 3 x
100
etc.
For symmetrical curves the y values can be calculated from one tangent only
for the entire curve.
Ht BVC = 857, 93-(150x5)/100 = 850,43m Ch BVC = 350-150 = 200m
Ht EVC = 857, 93-(150x3)/100 = 853,43m Ch EVC = 350+150 = 500m
g ( )
Position of highest point: x1 = 1 1
200 e
2
5(150)
=
= 187,50m
200 x3
2
g ( )
x2 = 2 2 = 112,50m
200 e
2
Check on  2 :
Tot
= 300, 00 OK
at Ch BVC + 187,5 = Ch 387,50m = Ch EVC – 112,50
PTO FOR CURVE DATA
Vertical Curves 112
SSUR221
SURVEYING II
Point
Chainage
x
Tangent Lev
y
Form Lev
BVC
200
0
850,43
0
850,43
220
20
851,43
-0,053
851,38
240
40
852,43
-0,213
852,22
260
60
853,43
-0,480
852,95
280
80
854,43
-0,853
853,58
300
100
855,43
-1,333
854,10
320
120
856,43
-1,920
854,51
340
140
857,43
-2,613
854,82
350
150
857,93
-3,000
854,93
360
160
858,43
-3,413
855,02
380
180
859,43
-4,320
855,11
387,50
187,50
859,81
-4,688
855,12
400
200
860,43
-5,333
855,10
420
220
861,43
-6,453
854,98
440
240
862,43
-7,680
854,75
460
260
863,43
-9,013
854,42
480
280
864,43
-10,453
853,98
500
300
865,43
-12,00
853,43
VPI
Highest Pt
EVC
EXAMPLE 2
A 4% downward gradient followed by a 3% downward gradient is joined by a 200m
symmetrical vertical curve. The elevation of the point of intersection of the gradients is
1000m above MSL. The curve begins at Chainage 0, 00. Determine elevations on the curve at
20m intervals.
G = -3-(-4) = +1%
e = 0,25m
Height of the BVC = 1004m
Height of the EVC = 997m
Vertical Curves 113
SSUR221
SURVEYING II
Point
Chainage
x
Tangent Lev
y
Form Lev
BVC
0
0
1004,00
0
1004,00
20
20
1003,20
+0,01
1003,21
40
40
1002,40
+0,04
1002,44
60
60
1001,60
+0,09
1001,69
80
80
1000,80
+0,16
1000,96
100
100
1000,00
+0,25
1000,25
120
120
999,20
+0,36
999,56
140
140
998,40
+0,49
998,89
160
160
997,60
+0,64
998,24
180
180
996,80
+0,81
997,61
200
200
996,00
+1,00
997,00
VPI
EVC
Vertical Curves 114
SSUR221
SURVEYING II
EXERCISES
QUESTION 1
The given data refer to a symmetrical vertical sag (valley) curve:
g 1 = -2,10%
g 2 = +3,20%
L = 260m
Chainage of the BVC = 2240,00m
Height of the VPI = 69,66m
1.1
Calculate the heights of the BVC and EVC
1.2
Calculate the height of each point on the curve at continuous chainages of 20m.
1.3
Calculate the chainage and height of the lowest point on the curve.
QUESTION 2
The following data applies to a symmetrical parabolic summit (crest) curve on a section of a
road:
g 1 = +4,00%
g 2 = -2,00%
L = 200m
Chainage of the VPI = 1160,00m
Height of the VPI = 230,00m
•
Calculate the heights of the BVC and EVC
•
Calculate the height of each point on the curve at continuous chainages of 20m.
•
Calculate the chainage and height of the highest point on the curve.
Vertical Curves 115
SSUR221
SURVEYING II
Chapter 10
COORDINATE CALCULATIONS
EMSH:
1.
Chapter VII (Certain sections - to be confirmed in class)
Direction and direction
EXAMPLE : Sin Method
P
G
D
Given data:
Coordinates of G and D, and oriented directions GP and DP
Co-ordinates: G
- 297,27
+2164,62
D
-1837,19
+4302,76
To calculate: Coordinates of intersection point P
Step 1:
Join GD
324.14.17
2634, 96m
Checked by polar 
Coordinate Calculation 116
SSUR221
SURVEYING II
Step 2: Calculate internal angles
324.14.17
246.45.45
211.02.22
G = 77.28.32
- 246.45.45
- 211.02.22
- 144.14.17
P = 35.43.23
G = 77.28.32
P = 35.43.23
D = 66.48.05
D = 66.48.05
Check:
Σ = 180.00.00
Step 3: Calculate sides GP and DP
GP
=
GD.SinD
SinP
DP
= 4148, 04m
=
GD.SinG
SinP
= 4405,55m
Step 4: Calculate Polars
Polar GP
246.45.45
4148, 04
Polar DP
P: -4108,81
+528,04
-6319,70
-6531,60
-6379,20
-6559,90
+4639,40
+4701,20
+5322,00
+5261,00
211.02.22
4405,55
P: -4108,81
+528,03
EXERCISES
1.
A
B
C
D
A and B are two pegs on one straight, C and D are on the other straight. Calculate
coordinates for the point of intersection P
2.
G and H lie on a straight, and C and D lie on a second straight of a proposed road.
C
D
G
H
-11274,24
- 9983,91
- 8424,21
- 9453,66
+3322,16
+5467,38
+3275,88
+2860,00
Calculate coordinates for the point of intersection P
EXAMPLE : Tan Method
Coordinate Calculation 117
SSUR221
SURVEYING II
C
B
A
Data: Coordinates of A and B, and oriented directions xAC and xBC
To be calculated:
Coordinates of C
It can be proved that:
XC − XA =
(YB − YA ) − (XB − XA ) tan xBC ……………………… (Ι )
XC − XB =
(YB − YA ) − (XB − XA ) tan xAC ……………………... (ΙΙ)
tan xAC − tan xBC
tan xAC − tan xBC
YC – YA = (XC – XA) tan xAC…………………………………….. (ΙΙΙ)
YC – YB = (XC – XB) tan xBC……………………………………. (IV )
Remembering the formulae above could be made easier by assigning numbers as below.
Store the calculated values in the memories of your calculator with the same numbers:
(YB – YA) = -2277.991 = 1
(XB – XA) = 1324.300=  2
tan xAZ = 0.922639 =  3
tan xBZ = 0.122298 =  4
and
YZ – YB
=  9
tan xAZ – tan xBZ = 0.800341 =  5
XZ – XA
=  6
XZ – XB
=  7
YZ – YA
=  8
The formulae above can now be written as follows:
Coordinate Calculation 118
SSUR221
SURVEYING II
(Ι) : XZ – XA = 1−  2 x  4 = -3048.638= 6
 5
(ΙΙ) : XZ – XB = 1−  2 x  3 = -4372.938= 7
 5
(ΙΙΙ) : YZ – YA = 6 x 3 = -2812.792= 8
(IV) : YZ – YB = 7 x 4 = -534.802= 9
Then XZ – XA + XA = XZ  6 + XA = -1548.638=XZ
and
XZ – XB + XB = XZ  7 + XB = -1548.638 =XZ (check)
Also
and
YZ – YA + YA = YZ  8 + YA = -1812.792=YZ
YZ – YB + YB = YZ  9 + YB = -1812.793 YZ (check)
Distance AZ =
 8 +  6  =4148.011m
Distance BZ =
 9 +  7  =4405.519m
2
2
2
2
Suggested form of calculation:
Coordinate Calculation 119
SSUR221
SURVEYING II
A
YA=+1000.000
XA=+1500.000
xAZ=222.41.45
B
YB= -1277.991
XB=+2824.300
xBZ=186.58.21
YB – YA =
-2277.991 = 1
XB – XA =
1324.300=  2
YZ – YA = -2812.792
=  8 (ΙΙΙ)
XZ – XA =
-3048.638= 6 (Ι )
tan xAZ =
0.922639 =  3
Distance AZ=4148.011m
YZ – YB = -534.799
 9 (IV)
XZ – XB =
-4372.936
 7 (ΙΙ)
tan xBZ
=0.122298 =  4
Distance BZ=4405.519m
 3–  4 =
0.800341 =  5
Z
8 + YA=-1812.792
6 + XA=-
=
9 + YB=-1812.793
1548.638
=
7 + XB=1548.638
-1812.793 = YZ
-1548.638= XZ
Examples – to be discussed in class
G
-297.27
+2164.,62
246.45.45
D
-1837.19
+4302.76
211.02.22
-1539.92(1)
+2138.14(2)
-3811.54(8)
-1636.58(6)
+2.3289........(3)
4148.04
-2271.62(9)
-3774.72(7)
+0.6017........(4)
4405.54
GP
-4108.81
+528.04
+1.7276........(5)
DP
-4108.81
+528.04
P
-4108.81
+528.04
Coordinate Calculation 120
SSUR221
SURVEYING II
L
-1416,54
+8715,92
340.16.41
R
-13661,99
+14967,62
71.08.33
-12245,45
+6251,70
-3332,44
+9295,92
-0,358484
9875,19m
+8913,01
+3044,22
+2,927846
9418,55m
-3,286330
P
-4748,98
+18011,84
= YP
= XP
Try exercises No 1, 2 and the Sin method example above, using the blank tables on the next
page.
Example
Coordinate Calculation 121
SSUR221
SURVEYING II
G
-297.27
+2164.62
xGC = 246.45.45
D
-1837.19
+4302.76
xDC = 211.02.22
-1539.92
+2138.14
-3811.54
-1636.58
+2.328963843
4148.04
-2271.62
-3774.72
+0.60179799
4405.54
-4108.81
+528.04
+1.72765852
-4108.81
+528.04
-4108.81
+528.04
P
1.
2.
Note: The directions used above could be from field observations or calculated using known
coordinate data (e.g. from the centre line of a road). If obtained from field
observations, more observations could be added from other known points, as well as
from the unknown point, and final coordinate values could be obtained by doing an
adjustment using all observations. Such an adjustment is beyond the scope of this
course.
2.
Distance and distance (Trilateration)
Coordinate Calculation 122
SSUR221
SURVEYING II
Trilateration is a method of fixing the position of an unknown point by measuring distances
to known points from the unknown. With modern electronic equipment this measurement can
be done very accurately. The method is not influenced by poor visibility and weather
conditions, as much as triangulation, and is also very quick.
EXAMPLE
Given data: Co-ordinates of A and B, and distances AC and BC, i.e. a and b
Required: to calculate Co-ordinates of C.
A
c
b
B
C
aa
Data:
A
B
D
E
-31 472,63
-34 683,83
-30 478,70
-33 483,62
+97 463,21
+95 678,01
+92 909,64
+91 537,91
Reduced distances:
a
b
CD
CE
2976,51m
4531,20m
2564,11m
1730,01m
Calculation sequence:
•
•
•
•
•
Join AB
Solve for  using Cosine Rule
Polar A(C)
Join B(C) and also from (C) to any other known point measured to
Adjustment and final C
1.
JOIN AB
240.55.45
3674,064
2.
Cosine Rule: Â = 40.53.21
Checked by polar 
xAC = 200.02.24
3.
POLAR A(C)
Coordinate Calculation 123
SSUR221
SURVEYING II
200.02.24
4531,20
4.
5.
(C): -33 025,36
+93 206,36
JOINS
(C) - B
326.08.19
2976,50m
(C) - D
96.38.45
2563,89m
(C) - E
195.21.29
1730,24m
DATA FOR CHECKING CORRECTNESS OF MEASUREMENTS
At (C)
Direction
Join Dist
Meas. Dist
Error
A
20.02.24
4531,20
4531,20
0,00
B
326.08.19
2976,50
2976,51
+0,01
D
96.38.45
2563,89
2564,11
+0,22
E
195.21.29
1730,24
1730,01
-0,23
Note: The data above could be used to obtain final coordinate values by doing an adjustment
using all measurements. Such an adjustment is beyond the scope of this course. If
possible, the two shortest distances, forming an angle of as close to 90° as possible at
the unknown point should be used to calculate provisional coordinates – (C) in the
example above.
EXERCISES
1.
Calculate coordinates for C from the following. Do all possible checks:
A
B
D
E
- 909,54
-1920,84
- 596,53
-1524,87
+2350,39
+1788,17
+ 916,32
+ 484,32
Measured distances
AC
BC
DC
EC
1427,01m
937,39m
807,51m
544,83m
Coordinate Calculation 124
SSUR221
2.
SURVEYING II
Calculate provisional coordinates for the point LION, from the points SPITS and
DIEP. Calculate the data necessary to check the correctness of the measurements.
Name
Y Co-ordinate X Co-ordinate
====================================
DIEP
+4445.72
+300496.91
KROM
- 944.32
+300027.27
PLAT
+1576.12
+298080.28
ROOI
+1486.47
+301109.24
SPITS
+3989.15
+298088.63
Triangulation Sketch: not to scale
PLAT
SPITS
LION
Measured distances:
KROM
LION
-
SPITS
DIEP
KROM
ROOI
PLAT
1895,72m
2061,06m
3660,31m
2038,03m
1753,16m
Coordinate Calculation 125
DIEP
ROOI
SSUR221
3.
SURVEYING II
Angular observations at the unknown point only (Resection)
Because observations are taken only at the unknown point, an orientation correction has to be
calculated before the triangle calculation can be done.
•
•
•
•
If three known points are visible and correctly identified from an unknown point,
by measuring three directions to these known points, the orientation correction, and
subsequently the coordinates of the unknown point can be determined.
A minimum of four existing known points is usually recommended to allow for
checks
The existing control points should be evenly distributed around the point
The existing control points observed should be identified in the field using a map
Conditions for a resection
• Good resection
►
Unknown point ‘P’ does not lie on the circle passing through the observed
points
• Poor resection
►
►
Unknown point ‘P’ lies on the circle passing through the observed points.
This configuration is referred to as a DANGER CIRCLE and makes resection
computation impossible as there is no unique solution
EXAMPLE (Collins’s q-point method)
Calculate the correction which must be applied to orient the observations at R. Use the
observations to D, A and B in that sequence.
@R
A
B
D
00.00.00
92.52.00
258.21.52
Co-ordinates
Resection 126
SSUR221
A
B
D
SURVEYING II
- 1850,24
- 4160,17
- 7286,45
Step 1:
+ 5860,90
+ 875,26
+ 4850,95
Join BD
321.49.13
5057, 64m
Checked by polar 
Step 2: Calculate angles at the unknown point (R)
92.52.00
180.00.00
258.21.52
360.00.00
92.52.00
- 00.00.00
-92.52.00
- 180.00.00
- 258.21.52
87.08.00
92.52.00
α = 87.08.00
β = 78.21.52
101.38.08
78.21.52
101.38.08
Check:
q = 180 – (α + β) = 14.30.08
Step 3: Calculate sides Bq and Dq
Bq
=
BD.Sin α
Sin q
= 20171, 56m
Dq
=
BD.Sin β
Sin q
= 19781, 79m
Step 4: Calculate directions Bq and Dq
Resection 127
Σ = 360.00.00
SSUR221
xBD
-β
xBq
:
:
=
SURVEYING II
321.49.13
- 78.21.52
243.27.21
xDB
+α
xDq
:
:
=
141.49.13
+87.08.00
228.57.13
Step 5: Calculate Polars
Polar Bq
Polar Dq
243.27.21
20171, 56
228.57.13
19781, 79
q: -22205,45 -8139,16
q: -22205,44 -8139,16
Step 6: Calculate orientation correction
Join qA:
55.28.49
Observed RA: 00.00.00
24704, 98m
Checked by polar 
Orientation correction = +55.28.49
This correction is now applied to all the observations at R, and a triangle calculation is
performed using the new oriented directions to calculate provisional coordinates for R. Once
again, more observations could be added from the unknown point, and final coordinate values
could be obtained by doing an adjustment using all observations. Such an adjustment is
beyond the scope of this course.
EXERCISES
1.
Calculate the orientation correction which must be applied to the following unoriented
observations at A:
Observations – not oriented @A
G
72.13.13
H
87.05.06
C
285.26.32
D
04.21.50
Co-ordinates: G
- 297, 27
+2164, 62
H
-1314, 02
+1728, 05
C
-3112, 13
+2184, 21
D
-1837, 19
+4302, 76
Use the rays from A to C, D and G in that sequence in your calculations.
2.
Below is an extract from a field book showing adjusted directions (See Chapter 4).
Calculate the orientation correction for the observations at Cable Hill. Use the rays to
A, B and D in that sequence in your calculations.
Resection 128
SSUR221
SURVEYING II
Station
CL
CR
Mean
Corr.
Adj. Dirn
A
80.59.28
261.02.14
81.00.51
0
81.00.51
B
154.41.21
334.44.05
154.42.43
-01
154.42.42
C
203.44.29
23.47.23
203.45.56
-01
203.45.55
E
206.30.30
26.33.16
206.31.53
-02
206.31.51
D
294.03.00
114.05.46
294.04.23
-02
294.04.21
RO
80.59.37
261.02.12
81.00.54
-03
@CABLE HILL
WG 21°
Constant
A
B
C
D
-60 000.00Y
- 9 814.24
- 9 625.55
-17 497.51
-22 812.27
+3 000 000.00 X
+ 28 685.07
+ 17 732.17
+ 19 614.24
+ 27 625.26
RESECTION: BLUNT’S METHOD
Resection 129
SSUR221
SURVEYING II
2

1
P


3
It can be proved that:
Tan x(P-Δ2)
=
(X1 − X 3 ) + (Y1 − Y2 )Cot α + (Y3 − Y2 )Cot β
(Y3 − Y1 ) + (X1 − X 2 )Cot α + (X 3 − X 2 )Cot β
=
d1
d2
Standard Form of Calculation: Write known points down in clockwise order with the
orienting ray in the middle. Store the calculated values in the memories of your calculator
with the numbers as indicated in curved brackets { }.
Resection 130
SSUR221
SURVEYING II
Δ1
Y1
x(P- Δ1)
X1
α
Y1 – Y2 {3}
X1 – X2 {4}
Y2
X2
Δ2
Cot α {5}
x(P- Δ2)
β
Y3 – Y2 {6}
X3 – X2 {7}
Y3
X3
Δ3
Cot β {8}
x(P- Δ3)
θ
Check:
Y3 – Y1 {1}
X1 – X3 {2}
{6} - {3} = {1}
{4} - {7} = {2}
α + β+ θ = 360º
The equation on the previous page now becomes:
d1
{2} + [{3}x{5}]+ [{6}x{8}]
=
= Tan x(P- Δ2)
d2
{1} + [{4}x{5}]+ [{7}x{8}]
d 
 x(P- Δ2) = Tan −1  1 
 d2 
Orientation correction = Calculated x(P- Δ2) – Observed x(P- Δ2)
Finding the correct quadrant for the calculated direction is as follows:
When
d1
d2
=
+
add 0º
+
=
+
add 180º
−
=
−
add 180º
−
−
add 360º (sometimes 180º)
+
If the directions obtained as explained above are out by 180º, the Tan method of calculating
the coordinates of intersecting lines will still yield correct coordinate values. A join can then
be done between the calculated point and Δ2 to check whether a 180º error still exists. This
=
Resection 131
SSUR221
SURVEYING II
will usually not be necessary since the surveyor usually knows the approximate range by
which he/ she was out of orientation when doing the observations in the field.
EXAMPLE (Chapter 5 page 74)
Calculate the correction which must be applied to orient the observations at R. Use the
observations to D, A and B in that sequence.
@R
A
B
D
Co-ordinates
00.00.00
92.52.00
258.21.52
D (1)
A
B
D
Y1 = - 7286,45
- 1850,24
- 4160,17
- 7286,45
+ 5860,90
+ 875,26
+ 4850,95
X1 = + 4850,95
258.21.52
α = 101.38.08
A (2)
Y1 – Y2 =
-5436,21 {3}
X1 – X2 =
-1009,95 {4}
Y2 = - 1850,24
X2 = + 5860,90
Cot α =
-0,205917312
{5}
00.00.00
Β = 92.52.00
B (3)
Y3 – Y2 =
-2309,93 {6}
X3 – X2
-4985,64 {7}
Y3 = - 4160,17
X3 = + 875,26
Cot β
-0,050074562
{8}
92.52.00
θ = 165.29.52
Check:
d1
d2
=
Y3 – Y1 =
+3126,28 {1}
X1 – X3
+3975,69 {2}
{6} - {3} = {1}
Checked
{4} - {7} = {2}
Checked
α + β+ θ = 360º
Checked
{3975,69} + [{-5436,21}x{-0,205917312}] + [{-2309,93}x{-0,050074562}]
{3126,28} + [{-1009,95}x{-0,205917312}] + [{-4985,64}x{-0,050074562}]
Resection 132
SSUR221
SURVEYING II
=
{3975,69} + [1119,409751] + [115,668733]
= Tan x(R- A)
{3126,28} + [207,9661893] + [249,6537393]
=
5210,768484
= 1,453938053
3583.899929
d 
x(R- A) = Tan −1  1 
 d2 
Observed x(R-A)
 Orientation correction
=
Tan −1 (1,453938053) + 0º
=
=
=
55.28.49
00.00.00
+55.28.49
Compare this answer with the answer obtained using the q-point method – Chapter 5 P. 75.
Once again, this correction is now applied to all the observations at R, and a triangle
calculation is performed using the new oriented directions to calculate provisional
coordinates for R.
Further Example: Resection using both methods.
@Jan
OBSERVED
Y
X
ΔZee
79.11.58
+45366,51
+78294,62
ΔKop
205.26.54
+37546,67
+74693,82
ΔTes
229.51.29
+38589,59
+77366,77
ΔVlei
291.47.35
+37907,52
+79544,18
ΔDuine
307.08.02
+36909,22
+81598,89
To determine the best arrangement of observed rays to use for the calculation of the
orientation correction, it is useful to determine the approximate position of the unknown point
– in this case Jan - on a triangulation sketch showing all the observed (known) points. One
way of doing this is to plot all the known points to scale on graph paper. The unoriented
observed rays are then drawn on transparent tracing paper. By moving the transparency over
the graph paper a unique fit should be found, indicating the position of the unknown point.
1.
Solution – q-point method - PTO
Resection 133
SSUR221
SURVEYING II
Resection 134
SSUR221
Step 1:
SURVEYING II
Join ΔDuine - ΔZee
111.20.26
9079, 87m
Checked by polar 
Step 2: Calculate angles at the unknown point (Jan)
205.26.54
307.08.02
79.11.58
360.00.00
126.14.56
- 79.11.58
-205.26.54
- 25.26.54
- 307.08.02
101.41.08
126.14.56
101.41.08
α = 53.45.04
+25.26.54
53.45.04
β = 78.18.52
78.18.52
Check:
q = 180 – (α + β) = 47.56.04
Resection 135
Σ = 360.00.00
SSUR221
SURVEYING II
Step 3: Calculate sides ΔZee-q and ΔDuine-q
(Note: Join distance = “S” in calculations below)
ΔDuine-q
=
S.Sin β
Sin q
=
11977, 28m
ΔZee-q
=
S.Sin α
Sin q
=
9863, 59m
Step 4: Calculate directions ΔDuine-q and ΔZee-q
x(ΔDuine – ΔZee) : 111.20.26
-α
:
- 53.45.04
x(ΔDuine-q) =
57.35.22
x(ΔZee – ΔDuine) : 291.20.26
+β
:
+78.18.52
x(ΔZee-q) =
09.39.18
Check : x(ΔDuine-q) - x(ΔZee-q) = 47.56.04 = q OK 
Step 5: Calculate Polars
Polar ΔDuine-q
Polar ΔZee-q
57.35.22
11977, 28
09.39.18
9863, 59
q: +47020,78 +88018,50
16349, 49m
Checked by polar 
q: +47020,79 +88018,50
Step 6: Calculate orientation correction
Join x(q-ΔKop):
215.24.49
Observed x(Jan-ΔKop):
205.26.54
Orientation correction = +09.57.55
@Jan
Observed Direction
Orientation
Correction
Oriented Direction
ΔZee
79.11.58
+09.57.55
89.09.52
ΔKop
205.26.54
215.24.49
ΔTes
229.51.29
239.49.24
ΔVlei
291.47.35
301.45.30
ΔDuine
307.08.02
317.05.27
The oriented directions can now be used to calculate adjusted coordinates for the
unknown point Jan. A calculation showing the adjusted point values appear at the end
of this section, but performing the adjustment routines falls outside the scope of this
module.
Resection 136
SSUR221
SURVEYING II
Solution – Blunt’s method
2.
ΔZee
+45366,51
+78294,62
79.11.58
α = 126.14.56
ΔKop
+7819,84 {3}
+3600,80 {4}
+37546,67
+74693,82
-0,733201 {5}
205.26.54
β = 101.41.08
ΔDuine
-637,45 {6}
+6905,07 {7}
+36909,22
+81598,89
-0,206827 {8}
307.08.02
θ = 132.03.56
Checked:
α + β+ θ = 360º
-8457,29 {1}
-3304,27 {2}
{6} - {3} = {1}
{4} - {7} = {2}
d1
- 8905,9388
=
= 0,711022
- 12525,5543
d2
d 
x(Jan-ΔKop) = Tan −1  1 
 d2 
=
Tan −1 (0,711022) + 180º
=
215.24.49
Observed x(Jan-ΔKop):
205.26.54
Orientation correction = +09.57.55
@Jan
Observed Direction
Orientation
Correction
Oriented Direction
ΔZee
79.11.58
+09.57.55
89.09.52
ΔKop
205.26.54
215.24.49
ΔTes
229.51.29
239.49.24
ΔVlei
291.47.35
301.45.30
ΔDuine
307.08.02
317.05.27
Again, the oriented directions can now be used to calculate adjusted coordinates for the
unknown point Jan.
Resection 137
SSUR221
SURVEYING II
FULL CALCULATION USING A SURVEY SOFTWARE PACKAGE.
The adjustment routines fall outside the scope of this module.
LEAST SQUARES ADJUSTMENT (SINGLE POINT FIX)
Resection Calculation for [Y, X] Determination
—————————————————————————————————————————————
Orientation Correction at [Jan] = +9:57:54
OBSERVED
FINAL
NAME
Y
X
BEARING
DIST
BEARING
DIST
————————————————————————————————————————————————————————————————————————————————————————
Zee
45366,5100
78294,6200
79:11:57
89:09:47
5315,3411
Kop
37546,6700
74693,8200 205:26:54
215:24:49
4322,9647
Tes
38589,5900
77366,7700 229:51:29
239:49:21
1691,3695
Vlei
37907,5200
79544,1800 291:47:34
301:45:22
2521,7290
Duine
36909,2200
81598,8900 307:08:01
317:06:04
4616,5691
Jan
40051,7360
78216,9817
======================================
(Standard Deviation of Unit Weight = 8,8")
ERROR ELLIPSE PARAMETERS
SIGU
SIGV
DIR(U)
————————————————————————
0,0777
0,1086
-28
LIST OF ADJUSTMENTS
POINT
BEARING
DIST.
NAME
(Secs)
(m)
——————————————————————————
Zee
-5,1
Kop
+1,4
Tes
-1,3
Vlei
-7,0
Duine
+8,9
BEARING CUTS
NAME
Y
X
—————————————————————————
Zee
-9,0533 0,1320
Kop
-0,0362 0,0509
Tes
0,0222 -0,0129
Vlei
-0,1634 -0,1012
Duine
0,2726 0,2934
Resection 138
SSUR221
3.
SURVEYING II
Angular observations at the unknown point only (Resection)
Because observations are taken only at the unknown point, an orientation correction has to be
calculated before the triangle calculation can be done.
•
•
•
•
If three known points are visible and correctly identified from an unknown point,
by measuring three directions to these known points, the orientation correction, and
subsequently the coordinates of the unknown point can be determined.
A minimum of four existing known points is usually recommended to allow for
checks
The existing control points should be evenly distributed around the point
The existing control points observed should be identified in the field using a map
Conditions for a resection
• Good resection
►
Unknown point ‘P’ does not lie on the circle passing through the observed
points
• Poor resection
►
►
Unknown point ‘P’ lies on the circle passing through the observed points.
This configuration is referred to as a DANGER CIRCLE and makes resection
computation impossible as there is no unique solution
EXAMPLE (Collins’s q-point method)
Calculate the correction which must be applied to orient the observations at R. Use the
observations to D, A and B in that sequence.
@R
A
B
D
00.00.00
92.52.00
258.21.52
Co-ordinates
Resection 126
SSUR221
A
B
D
SURVEYING II
- 1850,24
- 4160,17
- 7286,45
Step 1:
+ 5860,90
+ 875,26
+ 4850,95
Join BD
321.49.13
5057, 64m
Checked by polar 
Step 2: Calculate angles at the unknown point (R)
92.52.00
180.00.00
258.21.52
360.00.00
92.52.00
- 00.00.00
-92.52.00
- 180.00.00
- 258.21.52
87.08.00
92.52.00
α = 87.08.00
β = 78.21.52
101.38.08
78.21.52
101.38.08
Check:
q = 180 – (α + β) = 14.30.08
Step 3: Calculate sides Bq and Dq
Bq
=
BD.Sin α
Sin q
= 20171, 56m
Dq
=
BD.Sin β
Sin q
= 19781, 79m
Step 4: Calculate directions Bq and Dq
Resection 127
Σ = 360.00.00
SSUR221
xBD
-β
xBq
:
:
=
SURVEYING II
321.49.13
- 78.21.52
243.27.21
xDB
+α
xDq
:
:
=
141.49.13
+87.08.00
228.57.13
Step 5: Calculate Polars
Polar Bq
Polar Dq
243.27.21
20171, 56
228.57.13
19781, 79
q: -22205,45 -8139,16
q: -22205,44 -8139,16
Step 6: Calculate orientation correction
Join qA:
55.28.49
Observed RA: 00.00.00
24704, 98m
Checked by polar 
Orientation correction = +55.28.49
This correction is now applied to all the observations at R, and a triangle calculation is
performed using the new oriented directions to calculate provisional coordinates for R. Once
again, more observations could be added from the unknown point, and final coordinate values
could be obtained by doing an adjustment using all observations. Such an adjustment is
beyond the scope of this course.
EXERCISES
1.
Calculate the orientation correction which must be applied to the following unoriented
observations at A:
Observations – not oriented @A
G
72.13.13
H
87.05.06
C
285.26.32
D
04.21.50
Co-ordinates: G
- 297, 27
+2164, 62
H
-1314, 02
+1728, 05
C
-3112, 13
+2184, 21
D
-1837, 19
+4302, 76
Use the rays from A to C, D and G in that sequence in your calculations.
2.
Below is an extract from a field book showing adjusted directions (See Chapter 4).
Calculate the orientation correction for the observations at Cable Hill. Use the rays to
A, B and D in that sequence in your calculations.
Resection 128
SSUR221
SURVEYING II
Station
CL
CR
Mean
Corr.
Adj. Dirn
A
80.59.28
261.02.14
81.00.51
0
81.00.51
B
154.41.21
334.44.05
154.42.43
-01
154.42.42
C
203.44.29
23.47.23
203.45.56
-01
203.45.55
E
206.30.30
26.33.16
206.31.53
-02
206.31.51
D
294.03.00
114.05.46
294.04.23
-02
294.04.21
RO
80.59.37
261.02.12
81.00.54
-03
@CABLE HILL
WG 21°
Constant
A
B
C
D
-60 000.00Y
- 9 814.24
- 9 625.55
-17 497.51
-22 812.27
+3 000 000.00 X
+ 28 685.07
+ 17 732.17
+ 19 614.24
+ 27 625.26
RESECTION: BLUNT’S METHOD
Resection 129
SSUR221
SURVEYING II
2

1
P


3
It can be proved that:
Tan x(P-Δ2)
=
(X1 − X 3 ) + (Y1 − Y2 )Cot α + (Y3 − Y2 )Cot β
(Y3 − Y1 ) + (X1 − X 2 )Cot α + (X 3 − X 2 )Cot β
=
d1
d2
Standard Form of Calculation: Write known points down in clockwise order with the
orienting ray in the middle. Store the calculated values in the memories of your calculator
with the numbers as indicated in curved brackets { }.
Resection 130
SSUR221
SURVEYING II
Δ1
Y1
x(P- Δ1)
X1
α
Y1 – Y2 {3}
X1 – X2 {4}
Y2
X2
Δ2
Cot α {5}
x(P- Δ2)
β
Y3 – Y2 {6}
X3 – X2 {7}
Y3
X3
Δ3
Cot β {8}
x(P- Δ3)
θ
Check:
Y3 – Y1 {1}
X1 – X3 {2}
{6} - {3} = {1}
{4} - {7} = {2}
α + β+ θ = 360º
The equation on the previous page now becomes:
d1
{2} + [{3}x{5}]+ [{6}x{8}]
=
= Tan x(P- Δ2)
d2
{1} + [{4}x{5}]+ [{7}x{8}]
d 
 x(P- Δ2) = Tan −1  1 
 d2 
Orientation correction = Calculated x(P- Δ2) – Observed x(P- Δ2)
Finding the correct quadrant for the calculated direction is as follows:
When
d1
d2
=
+
add 0º
+
=
+
add 180º
−
=
−
add 180º
−
−
add 360º (sometimes 180º)
+
If the directions obtained as explained above are out by 180º, the Tan method of calculating
the coordinates of intersecting lines will still yield correct coordinate values. A join can then
be done between the calculated point and Δ2 to check whether a 180º error still exists. This
=
Resection 131
SSUR221
SURVEYING II
will usually not be necessary since the surveyor usually knows the approximate range by
which he/ she was out of orientation when doing the observations in the field.
EXAMPLE (Chapter 5 page 74)
Calculate the correction which must be applied to orient the observations at R. Use the
observations to D, A and B in that sequence.
@R
A
B
D
Co-ordinates
00.00.00
92.52.00
258.21.52
D (1)
A
B
D
Y1 = - 7286,45
- 1850,24
- 4160,17
- 7286,45
+ 5860,90
+ 875,26
+ 4850,95
X1 = + 4850,95
258.21.52
α = 101.38.08
A (2)
Y1 – Y2 =
-5436,21 {3}
X1 – X2 =
-1009,95 {4}
Y2 = - 1850,24
X2 = + 5860,90
Cot α =
-0,205917312
{5}
00.00.00
Β = 92.52.00
B (3)
Y3 – Y2 =
-2309,93 {6}
X3 – X2
-4985,64 {7}
Y3 = - 4160,17
X3 = + 875,26
Cot β
-0,050074562
{8}
92.52.00
θ = 165.29.52
Check:
d1
d2
=
Y3 – Y1 =
+3126,28 {1}
X1 – X3
+3975,69 {2}
{6} - {3} = {1}
Checked
{4} - {7} = {2}
Checked
α + β+ θ = 360º
Checked
{3975,69} + [{-5436,21}x{-0,205917312}] + [{-2309,93}x{-0,050074562}]
{3126,28} + [{-1009,95}x{-0,205917312}] + [{-4985,64}x{-0,050074562}]
Resection 132
SSUR221
SURVEYING II
=
{3975,69} + [1119,409751] + [115,668733]
= Tan x(R- A)
{3126,28} + [207,9661893] + [249,6537393]
=
5210,768484
= 1,453938053
3583.899929
d 
x(R- A) = Tan −1  1 
 d2 
Observed x(R-A)
 Orientation correction
=
Tan −1 (1,453938053) + 0º
=
=
=
55.28.49
00.00.00
+55.28.49
Compare this answer with the answer obtained using the q-point method – Chapter 5 P. 75.
Once again, this correction is now applied to all the observations at R, and a triangle
calculation is performed using the new oriented directions to calculate provisional
coordinates for R.
Further Example: Resection using both methods.
@Jan
OBSERVED
Y
X
ΔZee
79.11.58
+45366,51
+78294,62
ΔKop
205.26.54
+37546,67
+74693,82
ΔTes
229.51.29
+38589,59
+77366,77
ΔVlei
291.47.35
+37907,52
+79544,18
ΔDuine
307.08.02
+36909,22
+81598,89
To determine the best arrangement of observed rays to use for the calculation of the
orientation correction, it is useful to determine the approximate position of the unknown point
– in this case Jan - on a triangulation sketch showing all the observed (known) points. One
way of doing this is to plot all the known points to scale on graph paper. The unoriented
observed rays are then drawn on transparent tracing paper. By moving the transparency over
the graph paper a unique fit should be found, indicating the position of the unknown point.
1.
Solution – q-point method - PTO
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Resection 134
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Step 1:
SURVEYING II
Join ΔDuine - ΔZee
111.20.26
9079, 87m
Checked by polar 
Step 2: Calculate angles at the unknown point (Jan)
205.26.54
307.08.02
79.11.58
360.00.00
126.14.56
- 79.11.58
-205.26.54
- 25.26.54
- 307.08.02
101.41.08
126.14.56
101.41.08
α = 53.45.04
+25.26.54
53.45.04
β = 78.18.52
78.18.52
Check:
q = 180 – (α + β) = 47.56.04
Resection 135
Σ = 360.00.00
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Step 3: Calculate sides ΔZee-q and ΔDuine-q
(Note: Join distance = “S” in calculations below)
ΔDuine-q
=
S.Sin β
Sin q
=
11977, 28m
ΔZee-q
=
S.Sin α
Sin q
=
9863, 59m
Step 4: Calculate directions ΔDuine-q and ΔZee-q
x(ΔDuine – ΔZee) : 111.20.26
-α
:
- 53.45.04
x(ΔDuine-q) =
57.35.22
x(ΔZee – ΔDuine) : 291.20.26
+β
:
+78.18.52
x(ΔZee-q) =
09.39.18
Check : x(ΔDuine-q) - x(ΔZee-q) = 47.56.04 = q OK 
Step 5: Calculate Polars
Polar ΔDuine-q
Polar ΔZee-q
57.35.22
11977, 28
09.39.18
9863, 59
q: +47020,78 +88018,50
16349, 49m
Checked by polar 
q: +47020,79 +88018,50
Step 6: Calculate orientation correction
Join x(q-ΔKop):
215.24.49
Observed x(Jan-ΔKop):
205.26.54
Orientation correction = +09.57.55
@Jan
Observed Direction
Orientation
Correction
Oriented Direction
ΔZee
79.11.58
+09.57.55
89.09.52
ΔKop
205.26.54
215.24.49
ΔTes
229.51.29
239.49.24
ΔVlei
291.47.35
301.45.30
ΔDuine
307.08.02
317.05.27
The oriented directions can now be used to calculate adjusted coordinates for the
unknown point Jan. A calculation showing the adjusted point values appear at the end
of this section, but performing the adjustment routines falls outside the scope of this
module.
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Solution – Blunt’s method
2.
ΔZee
+45366,51
+78294,62
79.11.58
α = 126.14.56
ΔKop
+7819,84 {3}
+3600,80 {4}
+37546,67
+74693,82
-0,733201 {5}
205.26.54
β = 101.41.08
ΔDuine
-637,45 {6}
+6905,07 {7}
+36909,22
+81598,89
-0,206827 {8}
307.08.02
θ = 132.03.56
Checked:
α + β+ θ = 360º
-8457,29 {1}
-3304,27 {2}
{6} - {3} = {1}
{4} - {7} = {2}
d1
- 8905,9388
=
= 0,711022
- 12525,5543
d2
d 
x(Jan-ΔKop) = Tan −1  1 
 d2 
=
Tan −1 (0,711022) + 180º
=
215.24.49
Observed x(Jan-ΔKop):
205.26.54
Orientation correction = +09.57.55
@Jan
Observed Direction
Orientation
Correction
Oriented Direction
ΔZee
79.11.58
+09.57.55
89.09.52
ΔKop
205.26.54
215.24.49
ΔTes
229.51.29
239.49.24
ΔVlei
291.47.35
301.45.30
ΔDuine
307.08.02
317.05.27
Again, the oriented directions can now be used to calculate adjusted coordinates for the
unknown point Jan.
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FULL CALCULATION USING A SURVEY SOFTWARE PACKAGE.
The adjustment routines fall outside the scope of this module.
LEAST SQUARES ADJUSTMENT (SINGLE POINT FIX)
Resection Calculation for [Y, X] Determination
—————————————————————————————————————————————
Orientation Correction at [Jan] = +9:57:54
OBSERVED
FINAL
NAME
Y
X
BEARING
DIST
BEARING
DIST
————————————————————————————————————————————————————————————————————————————————————————
Zee
45366,5100
78294,6200
79:11:57
89:09:47
5315,3411
Kop
37546,6700
74693,8200 205:26:54
215:24:49
4322,9647
Tes
38589,5900
77366,7700 229:51:29
239:49:21
1691,3695
Vlei
37907,5200
79544,1800 291:47:34
301:45:22
2521,7290
Duine
36909,2200
81598,8900 307:08:01
317:06:04
4616,5691
Jan
40051,7360
78216,9817
======================================
(Standard Deviation of Unit Weight = 8,8")
ERROR ELLIPSE PARAMETERS
SIGU
SIGV
DIR(U)
————————————————————————
0,0777
0,1086
-28
LIST OF ADJUSTMENTS
POINT
BEARING
DIST.
NAME
(Secs)
(m)
——————————————————————————
Zee
-5,1
Kop
+1,4
Tes
-1,3
Vlei
-7,0
Duine
+8,9
BEARING CUTS
NAME
Y
X
—————————————————————————
Zee
-9,0533 0,1320
Kop
-0,0362 0,0509
Tes
0,0222 -0,0129
Vlei
-0,1634 -0,1012
Duine
0,2726 0,2934
Resection 138
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Chapter 11
INTRODUCTION TO THE GLOBAL POSITIONING SYSTEM
Source: GPS Basics, a publication used by kind permission of GeoSystems Africa
EMSH:
1.
Chapter XV (Sections 1 and 4)
What is GPS and what does it do?
GPS is the shortened form of NAVSTAR GPS. This is an acronym for NAVigation System
with Time And Ranging Global Positioning System.
GPS is a solution for one of man’s longest and most troublesome problems. It provides an
answer to the question .Where on earth am I?
One can imagine that this is an easy question to answer. You can easily locate yourself by
looking at objects that surround you and position yourself relative to them. But what if you
have no objects around you? What if you are in the middle of the desert or in the middle of
the ocean? For many centuries, this problem was solved by using the sun and stars to
navigate. Also, on land, surveyors and explorers used familiar reference points from which to
base their measurements or find their way.
These methods worked well within certain boundaries. Sun and stars cannot be seen when it
is cloudy. Also, even with the most precise measurements position cannot be determined very
accurately.
After the second world war, it became apparent to the U.S. Department of Defense that a
solution had to be found to the problem of accurate, absolute positioning. Several projects
and experiments ran during the next 25 years or so, including Transit, Timation, Loran,
Decca etc. All of these projects allowed positions to be determined but were limited in
accuracy or functionality.
At the beginning of the 1970s, a new project was proposed. GPS. This concept promised to
fulfil all the requirements of the US government, namely that one should be able to determine
one’s position accurately, at any point on the earth’s surface, at any time, in any weather
conditions.
GPS is a satellite-based system that uses a constellation of 24 satellites to give a user an
accurate position. It is important at this point to define ‘accurate’. To a hiker or soldier in the
desert, accurate means about 15m. To a ship in coastal waters, accurate means 5m. To a
surveyor, accurate means 1cm or less. GPS can be used to achieve all of these accuracies in
all of these applications, the difference being the type of GPS receiver used and the technique
employed.
GPS was originally designed for military use at any time anywhere on the surface of the
earth. Soon after the original proposals were made, it became clear that civilians could also
use GPS, and not only for personal positioning (as was intended for the military). The first
two major civilian applications to emerge were marine navigation and surveying. Nowadays
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applications range from in-car navigation through truck fleet management to automation of
construction machinery.
2.
System Overview
The total GPS configuration is comprised of three distinct segments:
•
•
•
2.1
The Space Segment. Satellites orbiting the earth.
The Control Segment. Stations positioned on the earth’s equator to control the
satellites
The User Segment. Anybody that receives and uses the GPS signal.
The Space Segment
The Space Segment is designed to consist of 24 satellites orbiting the earth at
approximately 20200km every 12 hours. At time of writing there are 26 operational
satellites orbiting the earth.
The space segment is so designed that there will be a minimum of 4 satellites visible
above a 15° cut-off angle at any point of the earth’s surface at any one time. Four
satellites are the minimum that must be visible for most applications. Experience
shows that there are usually at least 5 satellites visible above 15° for most of the time
and quite often there are 6 or 7 satellites visible.
Each GPS satellite has several very accurate atomic clocks on board. The clocks
operate at a fundamental frequency of 10.23MHz. This is used to generate the signals
that are broadcast from the satellite.
The satellites broadcast two carrier waves constantly. These carrier waves are in the
L-Band (used for radio), and travel to earth at the speed of light. These carrier waves
are derived from the fundamental frequency, generated by a very precise atomic
clock:
•
•
The L1 carrier is broadcast at 1575.42 MHz (10.23 x 154)
The L2 carrier is broadcast at 1227.60 MHz (10.23 x 120).
The L1 carrier then has two codes modulated upon it. The C/A Code or
Coarse/Acquisition Code is modulated at 1.023MHz (10.23/10) and the P-code or
Precision Code is modulated at 10.23MHz). The L2 carrier has just one code
modulated upon it. The L2 P-code is modulated at 10.23 MHz.
GPS receivers use the different codes to distinguish between satellites. The codes can
also be used as a basis for making pseudo range measurements and therefore calculate
a position.
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The Control Segment
The Control Segment consists of one master control station, 5 monitor stations and 4
ground antennas distributed amongst 5 locations roughly on the earth’s equator. The
Control Segment tracks the GPS satellites, updates their orbiting position and
calibrates and synchronises their clocks.
A further important function is to determine the orbit of each satellite and predict its
path for the following 24 hours. This information is uploaded to each satellite and
subsequently broadcast from it. This enables the GPS receiver to know where each
satellite can be expected to be found.
The satellite signals are read at Ascension, Diego Garcia and Kwajalein. The
measurements are then sent to the Master Control Station in Colorado Springs where
they are processed to determine any errors in each satellite. The information is then
sent back to the four monitor stations equipped with ground antennas and uploaded to
the satellites.
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The User Segment
The User Segment comprises of anyone using a GPS receiver to receive the GPS
signal and determine their position and/ or time. Typical applications within the user
segment are land navigation for hikers, vehicle location, surveying, marine
navigation, aerial navigation, machine control etc.
3.
How GPS works
There are several different methods for obtaining a position using GPS. The method used
depends on the accuracy required by the user and the type of GPS receiver available. Broadly
speaking, the techniques can be broken down into three basic classes:
Autonomous Navigation using a single stand-alone receiver. Used by hikers, ships
that are far out at sea and the military. Position Accuracy is better than 100m for
civilian users and about 20m for military users.
Differentially corrected positioning. More commonly known as DGPS, this gives an
accuracy of between 0.5-5m. Used for inshore marine navigation, GIS data
acquisition, precision farming etc.
3.1
Differential Phase position. Gives an accuracy of 0.5-20mm. Used for many
surveying tasks, machine control etc.
Simple Navigation
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This is the simplest technique employed by GPS receivers to instantaneously give a
position and height and/ or accurate time to a user. The accuracy obtained is better
than 100m (usually around the 30-50m mark) for civilian users and 5-15m for military
users. The reasons for the difference between civilian and military accuracies are
given later in this section. Receivers used for this type of operation are typically
small, highly portable handheld units with a low cost. More expensive handheld units
can give sub-5m accuracies.
3.1.1
Satellite ranging
All GPS positions are based on measuring the distance from the satellites to
the GPS receiver on the earth. This distance to each satellite can be determined
by the GPS receiver. The basic idea is that of trilateration, which many
surveyors use in their daily work. If you know the distance to three points
relative to your own position, you can determine your own position relative to
those three points. From the distance to one satellite we know that the position
of the receiver must be at some point on the surface of an imaginary sphere
which has it’s origin at the satellite. By intersecting three imaginary spheres
the receiver position can be determined.
The problem with GPS is that only pseudoranges and the time at which the
signal arrived at the receiver can be determined.
Thus there are four unknowns to determine; position (X, Y, Z) and time of
travel of the signal. Observing to four satellites produces four equations which
can be solved, enabling these unknowns to be determined.
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3.1.2
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Calculating the distance to the satellite
In order to calculate the distance to each satellite, one of Isaac Newton.s laws
of motion is used:
Distance = Velocity x Time
For instance, it is possible to calculate the distance a train has travelled if you
know the velocity it has been travelling and the time for which it has been
travelling at that velocity.
GPS requires the receiver to calculate the distance from the receiver to the
satellite.
The Velocity is the velocity of the radio signal. Radio waves travel at the
speed of light, 290,000 km per second (186,000 miles per second).
The Time is the time taken for the radio signal to travel from the satellite to
the GPS receiver. This is a little harder to calculate, since you need to know
when the radio signal left the satellite and when it reached the receiver.
3.1.3
Error Sources
Up until this point, it has been assumed that the position derived from GPS is
very accurate and free of error, but there are several sources of error that
degrade the GPS position from a theoretical few metres to tens of metres.
These error sources are:
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▪
▪
▪
▪
▪
▪
Ionospheric and atmospheric delays
Satellite and Receiver Clock Errors
Multipath
Dilution of Precision
Selective Availability (S/A)
Anti Spoofing (A-S)
1. Ionospheric and Atmospheric delays
As the satellite signal passes through the ionosphere, it can be slowed down,
the effect being similar to light refracted through a glass block. These
atmospheric delays can introduce an error in the range calculation as the
velocity of the signal is affected. (Light only has a constant velocity in a
vacuum).
The ionosphere does not introduce a constant delay on the signal. There are
several factors that influence the amount of delay caused by the ionosphere.
a. Satellite elevation. Signals from low elevation satellites will be affected
more than signals from higher elevation satellites. This is due to the increased
distance that the signal passes through the atmosphere.
b. The density of the ionosphere is affected by the sun. At night, there is
very little ionospheric influence. In the day, the sun increases the effect of the
ionosphere and slows down the signal.
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The amount by which the density of the ionosphere is increased varies with
solar cycles (sunspot activity). Sunspot activity peaks approximately every 11
years.
In addition to this, solar flares can also randomly occur and also have an effect
on the ionosphere. Ionospheric errors can be mitigated by using one of two
methods:
- The first method involves taking an average of the effect of the reduction in
velocity of light caused by the ionosphere. This correction factor can then be
applied to the range calculations. However, this relies on an average and
obviously this average condition does not occur all of the time. This method is
therefore not the optimum solution to Ionospheric Error mitigation.
- The second method involves using ‘dual-frequency’ GPS receivers. Such
receivers measure the L1 and the L2 frequencies of the GPS signal. It is
known that when a radio signal travels through the ionosphere it slows down
at a rate inversely proportional to it’s frequency. Hence, if the arrival times of
the two signals are compared, an accurate estimation of the delay can be made.
Note that this is only possible with dual frequency GPS receivers. Most
receivers built for navigation are single frequency.
c. Water Vapour also affects the GPS signal. Water vapour contained in the
atmosphere can also affect the GPS signal. This effect, which can result in a
position degradation can be reduced by using atmospheric models.
2. Satellite and Receiver clock errors
Even though the clocks in the satellite are very accurate (to about 3
nanoseconds), they do sometimes drift slightly and cause small errors,
affecting the accuracy of the position. The US Department of Defense
monitors the satellite clocks using the Control Segment (see section 2.2) and
can correct any drift that is found.
3. Multipath Errors
Multipath occurs when the receiver antenna is positioned close to a large
reflecting surface such as a lake or building. The satellite signal does not travel
directly to the antenna but hits the nearby object first and is reflected into the
antenna creating a false measurement.
Multipath can be reduced by use of special GPS antennas that incorporate a
ground plane (a circular, metallic disk about 50cm (2 feet) in diameter) that
prevent low elevation signals reaching the antenna.
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For highest accuracy, the preferred solution is use of a choke ring antenna. A
choke ring antenna has 4 or 5 concentric rings around the antenna that trap any
indirect signals.
Multipath only affects high accuracy, survey-type measurements. Simple
handheld navigation receivers do not employ such techniques.
4. Dilution of Precision
The Dilution of Precision (DOP) is a measure of the strength of satellite
geometry and is related to the spacing and position of the satellites in the sky.
The DOP can magnify the effect of satellite ranging errors.
The range to the satellite is affected by range errors previously described.
When the satellites are well spaced, the position can be determined as being
within the shaded area in the diagram and the possible error margin is small.
When the satellites are close together, the shaded area increases in size,
increasing the uncertainty of the position. Different types of Dilution of
Precision or DOP can be calculated depending on the dimension.
VDOP . Vertical Dilution of Precision. Gives accuracy degradation in vertical
direction.
HDOP . Horizontal Dilution of Precision. Gives accuracy degradation in
horizontal direction.
PDOP . Positional Dilution of Precision. Gives accuracy degradation in 3D
position.
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GDOP . Geometric Dilution of Precision. Gives accuracy degradation in 3D
position and time.
The most useful DOP to know is GDOP since this is a combination of all the
factors. Some receivers do however calculate PDOP or HDOP which do not
include the time component.
The best way of minimizing the effect of GDOP is to observe as many
satellites as possible. Remember however, that the signals from low elevation
satellites are generally influenced to a greater degree by most error sources.
As a general guide, when surveying with GPS it is best to observe satellites
that are 15° above the horizon. The most accurate positions will generally be
computed when the GDOP is low, (usually 8 or less).
5. Selective Availability (S/A)
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Selective Availability is a process applied by the U.S. Department of Defense
to the GPS signal. This is intended to deny civilian and hostile foreign powers
the full accuracy of GPS by subjecting the satellite clocks to a process known
as ‘dithering’ which alters their time slightly. Additionally, the ephemeris (or
path that the satellite will follow) is broadcast as being slightly different from
what it is in reality. The end result is degradation in position accuracy.
It is worth noting that S/A affects civilian users using a single GPS receiver to
obtain an autonomous position. Users of differential systems are not
significantly affected by S/A.
Currently, Selective Availability is switched off.
6. Anti-Spoofing (A-S)
Anti-Spoofing is similar to S/A in that it’s intention is to deny civilian and
hostile powers access to the P-code part of the GPS signal and hence force use
of the C/A code which has S/A applied to it.
Anti-Spoofing encrypts the P-code into a signal called the Y-code. Only users
with military GPS receivers (the US and it’s allies) can de-crypt the Y-code.
3.2
Differentially corrected positions (DGPS)
Many of the errors affecting the measurement of satellite range can be completely
eliminated or at least significantly reduced using differential measurement techniques.
DGPS allows the civilian user to increase position accuracy to 2-3m or less, making it
more useful for many civilian applications.
3.2.1
The Reference Receiver
The Reference receiver antenna is mounted on a previously measured point
with known coordinates. The receiver that is set at this point is known as the
Reference Receiver or Base Station.
The receiver is switched on and begins to track satellites. It can calculate an
autonomous position using the techniques mentioned in section 3.1. Because it
is on a known point, the reference receiver can estimate very precisely what
the ranges to the various satellites should be.
The reference receiver can therefore work out the difference between the
computed and measured range values. These differences are known as
corrections. The reference receiver is usually attached to a radio data link
which is used to broadcast these corrections.
3.2.2
The Rover receiver
The rover receiver is on the other end of these corrections. The rover receiver
has a radio data link attached to it that enables it to receive the range
corrections broadcast by the Reference Receiver.
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The Rover Receiver also calculates ranges to the satellites as described in
section 3.1. It then applies the range corrections received from the Reference.
This lets it calculate a much more accurate position than would be possible if
the uncorrected range measurements were used.
Using this technique, all of the error sources listed in section 3.1.3 are
minimized, hence the more accurate position.
It is also worthwhile to note that multiple Rover Receivers can receive
corrections from one single Reference.
3.2.3
Further details
DGPS has been explained in a very simple way in the preceding sections. In
real life, it is a little more complex than this.
One large consideration is the radio link. There are many types of radio link
that will broadcast over different ranges and frequencies. The performance of
the radio link depends on a range of factors including:
•
•
•
•
3.3
Frequency of the radio
Power of the radio
Type and .gain. of radio antenna
Antenna position
Differential Phase GPS and Ambiguity Resolution
Differential Phase GPS is used mainly in surveying and related industries to achieve
relative positioning accuracies of typically 5-50mm (0.25-2.5 in). The technique used
differs from previously described techniques and involves a lot of statistical analysis.
It is a differential technique which means that a minimum of two GPS receivers are
always used simultaneously. This is one of the similarities with the Differential Code
Correction method described in section 3.2.
The Reference Receiver is always positioned at a point with fixed or known
coordinates. The other receiver(s) are free to rove around. Thus they are known as
Rover Receivers. The baseline(s) between the Reference and Rover receiver(s) are
calculated.
The basic technique is still the same as with the techniques mentioned previously, measuring distances to four satellites and computing a position from those ranges.
The big difference comes in the way those ranges are calculated.
3.3.1
The Carrier Phase, C/A and P-codes
At this point, it is useful to define the various components of the GPS signal.
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Carrier Phase. The sine wave of the L1 or L2 signal that is created by the
satellite. The L1 carrier is generated at 1575.42MHz, the L2 carrier at 1227.6
MHz.
C/A code. The Coarse Acquisition code. Modulated on the L1 Carrier at
1.023MHz.
P-code. The precise code. Modulated on the L1 and L2 carriers at 10.23 MHz.
Refer also to the diagram in section 2.1.
What does modulation mean ?
The carrier waves are designed to carry the binary C/A and P-codes in a
process known as modulation. Modulation means the codes are superimposed
on the carrier wave. The codes are binary codes. This means they can only
have the values 1 or -1. Each time the value changes, there is a change in the
phase of the carrier.
3.3.2
Why use Carrier Phase?
The carrier phase is used because it can provide a much more accurate
measurement to the satellite than using the P-code or the C/A code. The L1
carrier wave has a wavelength of 19.4 cm. If you could measure the number of
wavelengths (whole and fractional parts) there are between the satellite and
receiver, you have a very accurate range to the satellite.
3.3.3
Double Differencing
The majority of the error incurred when making an autonomous position
comes from imperfections in the receiver and satellite clocks. One way of
bypassing this error is to use a technique known as Double Differencing.
If two GPS receivers make a measurement to two different satellites, the clock
offsets in the receivers and satellites cancel, removing any source of error that
they may contribute to the equation.
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3.3.4
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Ambiguity and Ambiguity Resolution
After removing the clock errors by double differencing, the whole number of
carrier wavelengths plus a fraction of a wavelength between the satellite and
receiver antenna can be determined. The problem is that there are many ‘sets’
of possible whole wavelengths to each observed satellite. Thus the solution is
ambiguous. Statistical processes can resolve this ambiguity and determine the
most probable solution.
4.
Surveying with GPS
Probably even more important to the surveyor or engineer than the theory behind GPS, are
the practicalities of the effective use of GPS.
Like any tool, GPS is only as good as it’s operator. Proper planning and preparation are
essential ingredients of a successful survey, as well as an awareness of the capabilities and
limitations of GPS.
Why use GPS?
GPS has numerous advantages over traditional surveying methods:
1. Intervisibility between points is not required.
2. Can be used at any time of the day or night and in any weather.
3. Produces results with very high geodetic accuracy.
4. More work can be accomplished in less time with fewer people.
Limitations
In order to operate with GPS it is important that the GPS Antenna has a clear view to at least
4 satellites. Sometimes, the satellite signals can be blocked by tall buildings, trees etc. Hence,
GPS cannot be used indoors. It is also difficult to use GPS in town centres or woodland.
Due to this limitation, it may prove more cost effective in some survey applications to use an
optical total station or to combine use of such an instrument with GPS.
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GPS Measuring Techniques
There are several measuring techniques that can be used by most GPS Survey
Receivers. The surveyor should choose the appropriate technique for the application.
Static - Used for long lines, geodetic networks, tectonic plate studies etc. Offers high
accuracy over long distances but is comparatively slow.
Rapid Static - Used for establishing local control networks, Network densification
etc. Offers high accuracy on baselines up to about 20km and is much faster than the
Static technique.
Kinematic - Used for detail surveys and measuring many points in quick succession.
Very efficient way of measuring many points that are close together. However, if
there are obstructions to the sky such as bridges, trees, tall buildings etc., and less than
4 satellites are tracked, the equipment must be reinitialized which can take 5-10
minutes. A processing technique known as On-the-Fly (OTF) has gone a long way to
minimise this restriction.
RTK - Real Time Kinematic uses a radio data link to transmit satellite data from the
Reference to the Rover. This enables coordinates to be calculated and displayed in
real time, as the survey is being carried out. Used for similar applications as
Kinematic. A very effective way for measuring detail as results are presented as work
is carried out.
This technique is however reliant upon a radio link, which is subject to interference
from other radio sources and also line of sight blockage.
4.1.1
Static Surveys
This was the first method to be developed for GPS surveying. It can be used
for measuring long baselines (usually 20km (16 miles) and over).
One receiver is placed on a point whose coordinates are known accurately in
WGS84. This is known as the Reference Receiver. The other receiver is
placed on the other end of the baseline and is known as the Rover.
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Data is then recorded at both stations simultaneously. It is important that data
is being recorded at the same rate at each station. The data collection rate may
be typically set to 15, 30 or 60 seconds.
The receivers have to collect data for a certain length of time. This time is
influenced by the length of the line, the number of satellites observed and the
satellite geometry (dilution of precision or DOP). As a rule of thumb, the
observation time is a minimum of 1 hour for a 20km line with 5 satellites and
a prevailing GDOP of 8. Longer lines require longer observation times.
Once enough data has been collected, the receivers can be switched off. The
Rover can then be moved to the next baseline and measurement can once
again commence.
It is very important to introduce redundancy into the network that is being
measured. This involves measuring points at least twice and creates safety
checks against problems that would otherwise go undetected.
A great increase in productivity can be realized with the addition of an extra
Rover receiver. Good coordination is required between the survey crews in
order to maximize the potential of having three receivers.
4.1.2
Rapid Static Surveys
In Rapid Static surveys, a Reference Point is chosen and one or more Rovers
operate with respect to it.
Typically, Rapid Static is used for densifying existing networks, establishing
control etc. When starting work in an area where no GPS surveying has
previously taken place, the first task is to observe a number of points, whose
coordinates are accurately known in the local system.
This will enable a transformation to be calculated and all hence, points
measured with GPS in that area can be easily converted into the local system.
As discussed in section 4.5, at least 4 known points on the perimeter of the
area of interest should be observed. The transformation calculated will then be
valid for the area enclosed by those points.
The Reference Receiver is usually set up at a known point and can be included
in the calculations of the transformation parameters. If no known point is
available, it can be set up anywhere within the network.
The Rover receiver(s) then visit each of the known points. The length of time
that the Rovers must observe for at each point is related to the baseline length
from the Reference and the GDOP.
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The data is recorded and post-processed back at the office. Checks should then
be carried out to ensure that no gross errors exist in the measurements. This
can done by measuring the points again at a different time of the day.
When working with two or more Rover receivers, an alternative is to ensure
that all rovers operate at each occupied point simultaneously. Thus allows data
from each station to be used as either Reference or Rover during post
processing and is the most efficient way to work, but also the most difficult to
synchronise.
Another way to build in redundancy is to set up two reference stations, and use
one rover to occupy the points.
4.1.3
Kinematic Surveys
The Kinematic technique is typically used for detail surveying, recording
trajectories etc., although with the advent of RTK its popularity is diminishing.
The technique involves a moving Rover whose position can be calculated
relative to the Reference.
Firstly, the Rover has to perform what is known as an initialization. This is
essentially the same as measuring a Rapid Static point and enables the post
processing software to resolve the ambiguity when back in the office. The
Reference and Rover are switched on and remain absolutely stationary for 520 minutes, collecting data. (The actual time depends on the baseline length
from the Reference and the number of satellites observed).
After this period, the Rover may then move freely. The user can record
positions at a predefined recording rate, can record distinct positions, or record
a combination of the two. This part of the measurement is commonly called
the kinematic chain.
A major point to watch during kinematic surveys is to avoid moving too close
to objects that could block the satellite signal from the Rover receiver. If at
any time, less than four satellites are tracked by the Rover receiver, you must
stop, move into a position where 4 or more satellites are tracked and perform
an initialization again before continuing.
Kinematic on the Fly - This is a variation of the Kinematic technique and
overcomes the requirement of initializing and subsequent re-initialization
when the number of observed satellites drops below four.
Kinematic on the Fly is a processing method that is applied to the
measurement during post-processing. At the start of measurement, the
operator can simply begin walking with the Rover receiver and record data. If
they walk under a tree and lose the satellites, upon emerging back into satellite
coverage, the system will automatically re-initialize.
4.1.4
RTK Surveys – NB!!!!!
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RTK stands for Real Time Kinematic. It is a Kinematic on the Fly survey
carried out in real time.
The Reference Station has a radio link attached and rebroadcasts the data it
receives from the satellites. The Rover also has a radio link and receives the
signal broadcast from the Reference. The Rover also receives satellite data
directly from the satellites via its own GPS Antenna. These two sets of data
can be processed together at the Rover to resolve the ambiguity and therefore
obtain a very accurate position relative to the Reference receiver.
Once the Reference Receiver has been set up and is broadcasting data through
the radio link, the Rover Receiver can be activated.
When it is tracking satellites and receiving data from the Reference, it can
begin the initialization process. This is similar to the initialization performed
in a post processed kinematic on the fly survey, the main difference being that
it is carried out in real-time.
Once the initialization is complete, the ambiguities are resolved and the Rover
can record point and coordinate data. At this time, baseline accuracies will be
in the 1 - 5cm range.
It is important to maintain contact with the Reference Receiver, otherwise the
Rover may lose the ambiguity. This results in a far less accurate position being
calculated. Additionally, problems may be encountered when surveying close
to obstructions such as tall buildings, trees etc. as the satellite signal may be
blocked.
RTK is quickly becoming the most common method of carrying out high
precision, high accuracy GPS surveys in small areas and can be used for
similar applications as a conventional total station. This includes detail
surveying, setting out etc.
The Radio Link - Most RTK GPS systems make use of small UHF radio
modems. Radio communication is the part of the RTK system that most
people experience difficulty with. It is worth considering the following
influencing factors when trying to optimize radio performance:
1.
2.
Power of the transmitting radio.
Generally speaking, the more power, the better the performance.
However, most countries legally restrict output power to 0.5 - 2W.
Height of transmitter antenna.
Radio communication can be affected by line of sight. The higher up
you can position the antenna, the less likely you are to get line of sight
problems. It will also increase the overall range of radio
communication. The same also applies to the receiving antenna.
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Other influencing factors affecting performance include the length of the cable
to radio antenna (longer cables mean higher losses) and the type of radio
antenna used.
EXERCISES
1.
Describe the following:
1.1
1.2
1.3
1.4
1.5
NAVSTAR
Ephemeris data
Selective Availability
Multipath
The GPS Space Segment
2.
Explain the difference between good and poor GDOP by means of two neat sketches
of the relevant geometric relationships between the receiver position and the positions
of the satellites.
3.
There are several measuring techniques that can be used by most GPS Survey
Receivers. The surveyor should choose the appropriate technique for the application.
Discuss, in point form, the measuring technique known as Real Time Kinematic.
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Chapter 11
INTRODUCTION TO THE GLOBAL POSITIONING SYSTEM
Source: GPS Basics, a publication used by kind permission of GeoSystems Africa
EMSH:
1.
Chapter XV (Sections 1 and 4)
What is GPS and what does it do?
GPS is the shortened form of NAVSTAR GPS. This is an acronym for NAVigation System
with Time And Ranging Global Positioning System.
GPS is a solution for one of man’s longest and most troublesome problems. It provides an
answer to the question .Where on earth am I?
One can imagine that this is an easy question to answer. You can easily locate yourself by
looking at objects that surround you and position yourself relative to them. But what if you
have no objects around you? What if you are in the middle of the desert or in the middle of
the ocean? For many centuries, this problem was solved by using the sun and stars to
navigate. Also, on land, surveyors and explorers used familiar reference points from which to
base their measurements or find their way.
These methods worked well within certain boundaries. Sun and stars cannot be seen when it
is cloudy. Also, even with the most precise measurements position cannot be determined very
accurately.
After the second world war, it became apparent to the U.S. Department of Defense that a
solution had to be found to the problem of accurate, absolute positioning. Several projects
and experiments ran during the next 25 years or so, including Transit, Timation, Loran,
Decca etc. All of these projects allowed positions to be determined but were limited in
accuracy or functionality.
At the beginning of the 1970s, a new project was proposed. GPS. This concept promised to
fulfil all the requirements of the US government, namely that one should be able to determine
one’s position accurately, at any point on the earth’s surface, at any time, in any weather
conditions.
GPS is a satellite-based system that uses a constellation of 24 satellites to give a user an
accurate position. It is important at this point to define ‘accurate’. To a hiker or soldier in the
desert, accurate means about 15m. To a ship in coastal waters, accurate means 5m. To a
surveyor, accurate means 1cm or less. GPS can be used to achieve all of these accuracies in
all of these applications, the difference being the type of GPS receiver used and the technique
employed.
GPS was originally designed for military use at any time anywhere on the surface of the
earth. Soon after the original proposals were made, it became clear that civilians could also
use GPS, and not only for personal positioning (as was intended for the military). The first
two major civilian applications to emerge were marine navigation and surveying. Nowadays
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applications range from in-car navigation through truck fleet management to automation of
construction machinery.
2.
System Overview
The total GPS configuration is comprised of three distinct segments:
•
•
•
2.2
The Space Segment. Satellites orbiting the earth.
The Control Segment. Stations positioned on the earth’s equator to control the
satellites
The User Segment. Anybody that receives and uses the GPS signal.
The Space Segment
The Space Segment is designed to consist of 24 satellites orbiting the earth at
approximately 20200km every 12 hours. At time of writing there are 26 operational
satellites orbiting the earth.
The space segment is so designed that there will be a minimum of 4 satellites visible
above a 15° cut-off angle at any point of the earth’s surface at any one time. Four
satellites are the minimum that must be visible for most applications. Experience
shows that there are usually at least 5 satellites visible above 15° for most of the time
and quite often there are 6 or 7 satellites visible.
Each GPS satellite has several very accurate atomic clocks on board. The clocks
operate at a fundamental frequency of 10.23MHz. This is used to generate the signals
that are broadcast from the satellite.
The satellites broadcast two carrier waves constantly. These carrier waves are in the
L-Band (used for radio), and travel to earth at the speed of light. These carrier waves
are derived from the fundamental frequency, generated by a very precise atomic
clock:
•
•
The L1 carrier is broadcast at 1575.42 MHz (10.23 x 154)
The L2 carrier is broadcast at 1227.60 MHz (10.23 x 120).
The L1 carrier then has two codes modulated upon it. The C/A Code or
Coarse/Acquisition Code is modulated at 1.023MHz (10.23/10) and the P-code or
Precision Code is modulated at 10.23MHz). The L2 carrier has just one code
modulated upon it. The L2 P-code is modulated at 10.23 MHz.
GPS receivers use the different codes to distinguish between satellites. The codes can
also be used as a basis for making pseudo range measurements and therefore calculate
a position.
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The Control Segment
The Control Segment consists of one master control station, 5 monitor stations and 4
ground antennas distributed amongst 5 locations roughly on the earth’s equator. The
Control Segment tracks the GPS satellites, updates their orbiting position and
calibrates and synchronises their clocks.
A further important function is to determine the orbit of each satellite and predict its
path for the following 24 hours. This information is uploaded to each satellite and
subsequently broadcast from it. This enables the GPS receiver to know where each
satellite can be expected to be found.
The satellite signals are read at Ascension, Diego Garcia and Kwajalein. The
measurements are then sent to the Master Control Station in Colorado Springs where
they are processed to determine any errors in each satellite. The information is then
sent back to the four monitor stations equipped with ground antennas and uploaded to
the satellites.
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The User Segment
The User Segment comprises of anyone using a GPS receiver to receive the GPS
signal and determine their position and/ or time. Typical applications within the user
segment are land navigation for hikers, vehicle location, surveying, marine
navigation, aerial navigation, machine control etc.
3.
How GPS works
There are several different methods for obtaining a position using GPS. The method used
depends on the accuracy required by the user and the type of GPS receiver available. Broadly
speaking, the techniques can be broken down into three basic classes:
Autonomous Navigation using a single stand-alone receiver. Used by hikers, ships
that are far out at sea and the military. Position Accuracy is better than 100m for
civilian users and about 20m for military users.
Differentially corrected positioning. More commonly known as DGPS, this gives an
accuracy of between 0.5-5m. Used for inshore marine navigation, GIS data
acquisition, precision farming etc.
3.1
Differential Phase position. Gives an accuracy of 0.5-20mm. Used for many
surveying tasks, machine control etc.
Simple Navigation
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This is the simplest technique employed by GPS receivers to instantaneously give a
position and height and/ or accurate time to a user. The accuracy obtained is better
than 100m (usually around the 30-50m mark) for civilian users and 5-15m for military
users. The reasons for the difference between civilian and military accuracies are
given later in this section. Receivers used for this type of operation are typically
small, highly portable handheld units with a low cost. More expensive handheld units
can give sub-5m accuracies.
3.1.1
Satellite ranging
All GPS positions are based on measuring the distance from the satellites to
the GPS receiver on the earth. This distance to each satellite can be determined
by the GPS receiver. The basic idea is that of trilateration, which many
surveyors use in their daily work. If you know the distance to three points
relative to your own position, you can determine your own position relative to
those three points. From the distance to one satellite we know that the position
of the receiver must be at some point on the surface of an imaginary sphere
which has it’s origin at the satellite. By intersecting three imaginary spheres
the receiver position can be determined.
The problem with GPS is that only pseudoranges and the time at which the
signal arrived at the receiver can be determined.
Thus there are four unknowns to determine; position (X, Y, Z) and time of
travel of the signal. Observing to four satellites produces four equations which
can be solved, enabling these unknowns to be determined.
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Calculating the distance to the satellite
In order to calculate the distance to each satellite, one of Isaac Newton.s laws
of motion is used:
Distance = Velocity x Time
For instance, it is possible to calculate the distance a train has travelled if you
know the velocity it has been travelling and the time for which it has been
travelling at that velocity.
GPS requires the receiver to calculate the distance from the receiver to the
satellite.
The Velocity is the velocity of the radio signal. Radio waves travel at the
speed of light, 290,000 km per second (186,000 miles per second).
The Time is the time taken for the radio signal to travel from the satellite to
the GPS receiver. This is a little harder to calculate, since you need to know
when the radio signal left the satellite and when it reached the receiver.
3.1.5
Error Sources
Up until this point, it has been assumed that the position derived from GPS is
very accurate and free of error, but there are several sources of error that
degrade the GPS position from a theoretical few metres to tens of metres.
These error sources are:
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▪
▪
▪
▪
▪
▪
Ionospheric and atmospheric delays
Satellite and Receiver Clock Errors
Multipath
Dilution of Precision
Selective Availability (S/A)
Anti Spoofing (A-S)
2. Ionospheric and Atmospheric delays
As the satellite signal passes through the ionosphere, it can be slowed down,
the effect being similar to light refracted through a glass block. These
atmospheric delays can introduce an error in the range calculation as the
velocity of the signal is affected. (Light only has a constant velocity in a
vacuum).
The ionosphere does not introduce a constant delay on the signal. There are
several factors that influence the amount of delay caused by the ionosphere.
a. Satellite elevation. Signals from low elevation satellites will be affected
more than signals from higher elevation satellites. This is due to the increased
distance that the signal passes through the atmosphere.
b. The density of the ionosphere is affected by the sun. At night, there is
very little ionospheric influence. In the day, the sun increases the effect of the
ionosphere and slows down the signal.
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The amount by which the density of the ionosphere is increased varies with
solar cycles (sunspot activity). Sunspot activity peaks approximately every 11
years.
In addition to this, solar flares can also randomly occur and also have an effect
on the ionosphere. Ionospheric errors can be mitigated by using one of two
methods:
- The first method involves taking an average of the effect of the reduction in
velocity of light caused by the ionosphere. This correction factor can then be
applied to the range calculations. However, this relies on an average and
obviously this average condition does not occur all of the time. This method is
therefore not the optimum solution to Ionospheric Error mitigation.
- The second method involves using ‘dual-frequency’ GPS receivers. Such
receivers measure the L1 and the L2 frequencies of the GPS signal. It is
known that when a radio signal travels through the ionosphere it slows down
at a rate inversely proportional to it’s frequency. Hence, if the arrival times of
the two signals are compared, an accurate estimation of the delay can be made.
Note that this is only possible with dual frequency GPS receivers. Most
receivers built for navigation are single frequency.
c. Water Vapour also affects the GPS signal. Water vapour contained in the
atmosphere can also affect the GPS signal. This effect, which can result in a
position degradation can be reduced by using atmospheric models.
2. Satellite and Receiver clock errors
Even though the clocks in the satellite are very accurate (to about 3
nanoseconds), they do sometimes drift slightly and cause small errors,
affecting the accuracy of the position. The US Department of Defense
monitors the satellite clocks using the Control Segment (see section 2.2) and
can correct any drift that is found.
3. Multipath Errors
Multipath occurs when the receiver antenna is positioned close to a large
reflecting surface such as a lake or building. The satellite signal does not travel
directly to the antenna but hits the nearby object first and is reflected into the
antenna creating a false measurement.
Multipath can be reduced by use of special GPS antennas that incorporate a
ground plane (a circular, metallic disk about 50cm (2 feet) in diameter) that
prevent low elevation signals reaching the antenna.
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For highest accuracy, the preferred solution is use of a choke ring antenna. A
choke ring antenna has 4 or 5 concentric rings around the antenna that trap any
indirect signals.
Multipath only affects high accuracy, survey-type measurements. Simple
handheld navigation receivers do not employ such techniques.
4. Dilution of Precision
The Dilution of Precision (DOP) is a measure of the strength of satellite
geometry and is related to the spacing and position of the satellites in the sky.
The DOP can magnify the effect of satellite ranging errors.
The range to the satellite is affected by range errors previously described.
When the satellites are well spaced, the position can be determined as being
within the shaded area in the diagram and the possible error margin is small.
When the satellites are close together, the shaded area increases in size,
increasing the uncertainty of the position. Different types of Dilution of
Precision or DOP can be calculated depending on the dimension.
VDOP . Vertical Dilution of Precision. Gives accuracy degradation in vertical
direction.
HDOP . Horizontal Dilution of Precision. Gives accuracy degradation in
horizontal direction.
PDOP . Positional Dilution of Precision. Gives accuracy degradation in 3D
position.
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GDOP . Geometric Dilution of Precision. Gives accuracy degradation in 3D
position and time.
The most useful DOP to know is GDOP since this is a combination of all the
factors. Some receivers do however calculate PDOP or HDOP which do not
include the time component.
The best way of minimizing the effect of GDOP is to observe as many
satellites as possible. Remember however, that the signals from low elevation
satellites are generally influenced to a greater degree by most error sources.
As a general guide, when surveying with GPS it is best to observe satellites
that are 15° above the horizon. The most accurate positions will generally be
computed when the GDOP is low, (usually 8 or less).
5. Selective Availability (S/A)
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Selective Availability is a process applied by the U.S. Department of Defense
to the GPS signal. This is intended to deny civilian and hostile foreign powers
the full accuracy of GPS by subjecting the satellite clocks to a process known
as ‘dithering’ which alters their time slightly. Additionally, the ephemeris (or
path that the satellite will follow) is broadcast as being slightly different from
what it is in reality. The end result is degradation in position accuracy.
It is worth noting that S/A affects civilian users using a single GPS receiver to
obtain an autonomous position. Users of differential systems are not
significantly affected by S/A.
Currently, Selective Availability is switched off.
6. Anti-Spoofing (A-S)
Anti-Spoofing is similar to S/A in that it’s intention is to deny civilian and
hostile powers access to the P-code part of the GPS signal and hence force use
of the C/A code which has S/A applied to it.
Anti-Spoofing encrypts the P-code into a signal called the Y-code. Only users
with military GPS receivers (the US and it’s allies) can de-crypt the Y-code.
3.2
Differentially corrected positions (DGPS)
Many of the errors affecting the measurement of satellite range can be completely
eliminated or at least significantly reduced using differential measurement techniques.
DGPS allows the civilian user to increase position accuracy to 2-3m or less, making it
more useful for many civilian applications.
3.2.1
The Reference Receiver
The Reference receiver antenna is mounted on a previously measured point
with known coordinates. The receiver that is set at this point is known as the
Reference Receiver or Base Station.
The receiver is switched on and begins to track satellites. It can calculate an
autonomous position using the techniques mentioned in section 3.1. Because it
is on a known point, the reference receiver can estimate very precisely what
the ranges to the various satellites should be.
The reference receiver can therefore work out the difference between the
computed and measured range values. These differences are known as
corrections. The reference receiver is usually attached to a radio data link
which is used to broadcast these corrections.
3.2.4
The Rover receiver
The rover receiver is on the other end of these corrections. The rover receiver
has a radio data link attached to it that enables it to receive the range
corrections broadcast by the Reference Receiver.
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The Rover Receiver also calculates ranges to the satellites as described in
section 3.1. It then applies the range corrections received from the Reference.
This lets it calculate a much more accurate position than would be possible if
the uncorrected range measurements were used.
Using this technique, all of the error sources listed in section 3.1.3 are
minimized, hence the more accurate position.
It is also worthwhile to note that multiple Rover Receivers can receive
corrections from one single Reference.
3.2.5
Further details
DGPS has been explained in a very simple way in the preceding sections. In
real life, it is a little more complex than this.
One large consideration is the radio link. There are many types of radio link
that will broadcast over different ranges and frequencies. The performance of
the radio link depends on a range of factors including:
•
•
•
•
3.3
Frequency of the radio
Power of the radio
Type and .gain. of radio antenna
Antenna position
Differential Phase GPS and Ambiguity Resolution
Differential Phase GPS is used mainly in surveying and related industries to achieve
relative positioning accuracies of typically 5-50mm (0.25-2.5 in). The technique used
differs from previously described techniques and involves a lot of statistical analysis.
It is a differential technique which means that a minimum of two GPS receivers are
always used simultaneously. This is one of the similarities with the Differential Code
Correction method described in section 3.2.
The Reference Receiver is always positioned at a point with fixed or known
coordinates. The other receiver(s) are free to rove around. Thus they are known as
Rover Receivers. The baseline(s) between the Reference and Rover receiver(s) are
calculated.
The basic technique is still the same as with the techniques mentioned previously, measuring distances to four satellites and computing a position from those ranges.
The big difference comes in the way those ranges are calculated.
3.3.1
The Carrier Phase, C/A and P-codes
At this point, it is useful to define the various components of the GPS signal.
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Carrier Phase. The sine wave of the L1 or L2 signal that is created by the
satellite. The L1 carrier is generated at 1575.42MHz, the L2 carrier at 1227.6
MHz.
C/A code. The Coarse Acquisition code. Modulated on the L1 Carrier at
1.023MHz.
P-code. The precise code. Modulated on the L1 and L2 carriers at 10.23 MHz.
Refer also to the diagram in section 2.1.
What does modulation mean ?
The carrier waves are designed to carry the binary C/A and P-codes in a
process known as modulation. Modulation means the codes are superimposed
on the carrier wave. The codes are binary codes. This means they can only
have the values 1 or -1. Each time the value changes, there is a change in the
phase of the carrier.
3.3.2
Why use Carrier Phase?
The carrier phase is used because it can provide a much more accurate
measurement to the satellite than using the P-code or the C/A code. The L1
carrier wave has a wavelength of 19.4 cm. If you could measure the number of
wavelengths (whole and fractional parts) there are between the satellite and
receiver, you have a very accurate range to the satellite.
3.3.3
Double Differencing
The majority of the error incurred when making an autonomous position
comes from imperfections in the receiver and satellite clocks. One way of
bypassing this error is to use a technique known as Double Differencing.
If two GPS receivers make a measurement to two different satellites, the clock
offsets in the receivers and satellites cancel, removing any source of error that
they may contribute to the equation.
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Ambiguity and Ambiguity Resolution
After removing the clock errors by double differencing, the whole number of
carrier wavelengths plus a fraction of a wavelength between the satellite and
receiver antenna can be determined. The problem is that there are many ‘sets’
of possible whole wavelengths to each observed satellite. Thus the solution is
ambiguous. Statistical processes can resolve this ambiguity and determine the
most probable solution.
4.
Surveying with GPS
Probably even more important to the surveyor or engineer than the theory behind GPS, are
the practicalities of the effective use of GPS.
Like any tool, GPS is only as good as it’s operator. Proper planning and preparation are
essential ingredients of a successful survey, as well as an awareness of the capabilities and
limitations of GPS.
Why use GPS?
GPS has numerous advantages over traditional surveying methods:
1. Intervisibility between points is not required.
2. Can be used at any time of the day or night and in any weather.
3. Produces results with very high geodetic accuracy.
4. More work can be accomplished in less time with fewer people.
Limitations
In order to operate with GPS it is important that the GPS Antenna has a clear view to at least
4 satellites. Sometimes, the satellite signals can be blocked by tall buildings, trees etc. Hence,
GPS cannot be used indoors. It is also difficult to use GPS in town centres or woodland.
Due to this limitation, it may prove more cost effective in some survey applications to use an
optical total station or to combine use of such an instrument with GPS.
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GPS Measuring Techniques
There are several measuring techniques that can be used by most GPS Survey
Receivers. The surveyor should choose the appropriate technique for the application.
Static - Used for long lines, geodetic networks, tectonic plate studies etc. Offers high
accuracy over long distances but is comparatively slow.
Rapid Static - Used for establishing local control networks, Network densification
etc. Offers high accuracy on baselines up to about 20km and is much faster than the
Static technique.
Kinematic - Used for detail surveys and measuring many points in quick succession.
Very efficient way of measuring many points that are close together. However, if
there are obstructions to the sky such as bridges, trees, tall buildings etc., and less than
4 satellites are tracked, the equipment must be reinitialized which can take 5-10
minutes. A processing technique known as On-the-Fly (OTF) has gone a long way to
minimise this restriction.
RTK - Real Time Kinematic uses a radio data link to transmit satellite data from the
Reference to the Rover. This enables coordinates to be calculated and displayed in
real time, as the survey is being carried out. Used for similar applications as
Kinematic. A very effective way for measuring detail as results are presented as work
is carried out.
This technique is however reliant upon a radio link, which is subject to interference
from other radio sources and also line of sight blockage.
4.1.1
Static Surveys
This was the first method to be developed for GPS surveying. It can be used
for measuring long baselines (usually 20km (16 miles) and over).
One receiver is placed on a point whose coordinates are known accurately in
WGS84. This is known as the Reference Receiver. The other receiver is
placed on the other end of the baseline and is known as the Rover.
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Data is then recorded at both stations simultaneously. It is important that data
is being recorded at the same rate at each station. The data collection rate may
be typically set to 15, 30 or 60 seconds.
The receivers have to collect data for a certain length of time. This time is
influenced by the length of the line, the number of satellites observed and the
satellite geometry (dilution of precision or DOP). As a rule of thumb, the
observation time is a minimum of 1 hour for a 20km line with 5 satellites and
a prevailing GDOP of 8. Longer lines require longer observation times.
Once enough data has been collected, the receivers can be switched off. The
Rover can then be moved to the next baseline and measurement can once
again commence.
It is very important to introduce redundancy into the network that is being
measured. This involves measuring points at least twice and creates safety
checks against problems that would otherwise go undetected.
A great increase in productivity can be realized with the addition of an extra
Rover receiver. Good coordination is required between the survey crews in
order to maximize the potential of having three receivers.
4.1.2
Rapid Static Surveys
In Rapid Static surveys, a Reference Point is chosen and one or more Rovers
operate with respect to it.
Typically, Rapid Static is used for densifying existing networks, establishing
control etc. When starting work in an area where no GPS surveying has
previously taken place, the first task is to observe a number of points, whose
coordinates are accurately known in the local system.
This will enable a transformation to be calculated and all hence, points
measured with GPS in that area can be easily converted into the local system.
As discussed in section 4.5, at least 4 known points on the perimeter of the
area of interest should be observed. The transformation calculated will then be
valid for the area enclosed by those points.
The Reference Receiver is usually set up at a known point and can be included
in the calculations of the transformation parameters. If no known point is
available, it can be set up anywhere within the network.
The Rover receiver(s) then visit each of the known points. The length of time
that the Rovers must observe for at each point is related to the baseline length
from the Reference and the GDOP.
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The data is recorded and post-processed back at the office. Checks should then
be carried out to ensure that no gross errors exist in the measurements. This
can done by measuring the points again at a different time of the day.
When working with two or more Rover receivers, an alternative is to ensure
that all rovers operate at each occupied point simultaneously. Thus allows data
from each station to be used as either Reference or Rover during post
processing and is the most efficient way to work, but also the most difficult to
synchronise.
Another way to build in redundancy is to set up two reference stations, and use
one rover to occupy the points.
4.1.3
Kinematic Surveys
The Kinematic technique is typically used for detail surveying, recording
trajectories etc., although with the advent of RTK its popularity is diminishing.
The technique involves a moving Rover whose position can be calculated
relative to the Reference.
Firstly, the Rover has to perform what is known as an initialization. This is
essentially the same as measuring a Rapid Static point and enables the post
processing software to resolve the ambiguity when back in the office. The
Reference and Rover are switched on and remain absolutely stationary for 520 minutes, collecting data. (The actual time depends on the baseline length
from the Reference and the number of satellites observed).
After this period, the Rover may then move freely. The user can record
positions at a predefined recording rate, can record distinct positions, or record
a combination of the two. This part of the measurement is commonly called
the kinematic chain.
A major point to watch during kinematic surveys is to avoid moving too close
to objects that could block the satellite signal from the Rover receiver. If at
any time, less than four satellites are tracked by the Rover receiver, you must
stop, move into a position where 4 or more satellites are tracked and perform
an initialization again before continuing.
Kinematic on the Fly - This is a variation of the Kinematic technique and
overcomes the requirement of initializing and subsequent re-initialization
when the number of observed satellites drops below four.
Kinematic on the Fly is a processing method that is applied to the
measurement during post-processing. At the start of measurement, the
operator can simply begin walking with the Rover receiver and record data. If
they walk under a tree and lose the satellites, upon emerging back into satellite
coverage, the system will automatically re-initialize.
4.1.4
RTK Surveys – NB!!!!!
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RTK stands for Real Time Kinematic. It is a Kinematic on the Fly survey
carried out in real time.
The Reference Station has a radio link attached and rebroadcasts the data it
receives from the satellites. The Rover also has a radio link and receives the
signal broadcast from the Reference. The Rover also receives satellite data
directly from the satellites via its own GPS Antenna. These two sets of data
can be processed together at the Rover to resolve the ambiguity and therefore
obtain a very accurate position relative to the Reference receiver.
Once the Reference Receiver has been set up and is broadcasting data through
the radio link, the Rover Receiver can be activated.
When it is tracking satellites and receiving data from the Reference, it can
begin the initialization process. This is similar to the initialization performed
in a post processed kinematic on the fly survey, the main difference being that
it is carried out in real-time.
Once the initialization is complete, the ambiguities are resolved and the Rover
can record point and coordinate data. At this time, baseline accuracies will be
in the 1 - 5cm range.
It is important to maintain contact with the Reference Receiver, otherwise the
Rover may lose the ambiguity. This results in a far less accurate position being
calculated. Additionally, problems may be encountered when surveying close
to obstructions such as tall buildings, trees etc. as the satellite signal may be
blocked.
RTK is quickly becoming the most common method of carrying out high
precision, high accuracy GPS surveys in small areas and can be used for
similar applications as a conventional total station. This includes detail
surveying, setting out etc.
The Radio Link - Most RTK GPS systems make use of small UHF radio
modems. Radio communication is the part of the RTK system that most
people experience difficulty with. It is worth considering the following
influencing factors when trying to optimize radio performance:
1.
2.
Power of the transmitting radio.
Generally speaking, the more power, the better the performance.
However, most countries legally restrict output power to 0.5 - 2W.
Height of transmitter antenna.
Radio communication can be affected by line of sight. The higher up
you can position the antenna, the less likely you are to get line of sight
problems. It will also increase the overall range of radio
communication. The same also applies to the receiving antenna.
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Other influencing factors affecting performance include the length of the cable
to radio antenna (longer cables mean higher losses) and the type of radio
antenna used.
EXERCISES
1.
Describe the following:
1.1
1.2
1.3
1.4
1.5
NAVSTAR
Ephemeris data
Selective Availability
Multipath
The GPS Space Segment
2.
Explain the difference between good and poor GDOP by means of two neat sketches
of the relevant geometric relationships between the receiver position and the positions
of the satellites.
3.
There are several measuring techniques that can be used by most GPS Survey
Receivers. The surveyor should choose the appropriate technique for the application.
Discuss, in point form, the measuring technique known as Real Time Kinematic.
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