Engineering surveying II

TACHEOMETRY

This is a surveying method in which both horizontal and vertical distances are determined by taking
angular observations with a tacheometer. Tacheometry is a Greek word meaning rapid/speedy
measurements.

Therefore here measurements of distances are made so fast using optical properties of a telescope
(ODM – Optical distance measurement) and not chaining (Direct measurement).

Advantages of tacheometry

 Measurements are taken above the ground and hence the poor surface measuring conditions
don’t affect it. i.e. can be adopted for rough and difficult terrains where5 direct leveling and
chaining not possible.
 Measurements are carried out in shorter time than that required by surface taping (chain
surveying)
 Has high accuracy than normal ground taping. i.e. a reasonable contour map can be prepared for
the investigated works.

Instruments employed in Tacheometry

There are basically two instruments used here, namely; tacheometer and a leveling staff/or stadia rod.

1) Tacheometer – is a transit theodolite fitted with stadia diaphragm and an anallatic lens.

Where the anallatic lens is an additional lens provided in the telescope between the objective glass and
diaphragm which reduces all observations to the centre of the instrument and thus eliminating the additive
constant k (reducing the additive constant to zero).
NB. All modern internal focusing telescopes, although not strictly anallactic may be regarded as so.
Advantages

 Reduces the additive constant to zero; therefore calculations for heights and distances are
simplified say ; .
 It’s free from moisture or dust.
 Loss of light is compensated by the use of slightly larger objective lens.

Disadvantages

 Brightness of the image is lost

 It can’t be easily cleaned

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 It’s also a potential source of errors.

2) Leveling staff /or stadia rod – is a stave normally 4 – 5m long with graduations with a possibility
of taking about minimum reading. This leveling stave has provisions for folding
and it’s suitable for short distance surveys.
Where as Stadia rod is suited for longer sights, has bold and clear graduations also taken to a
minimum reading of 0.001m. It’s also 4m long and can be folded or telescopic.

Principles of tacheometry

It is based on the properties of an isosceles triangle as shown below, where angle is the instrument
angle called the parallactic angle.

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⁄
⁄ ⁄ ( ⁄ )

( ) ⁄

The distance between the instrument and the staff is directly proportional to the staff intercept S.

The parallatic angle may either be fixed or variable depending on the type of theodolite (tacheometer)
used. This therefore defines the five different systems of Tacheometry.

Systems of tacheometry

 Stadia system -
 Tangential system -
 Subtense system –
 Optical wedge –

Tangential system

In this system, there are two (2) telescopic pointing is made for two readings to be taken.

It is the most easily understood but the least accurate.

Procedure

 Set up a theodolite at A and take horizontal reading C at B

 Sight the staff at an angle of elevation & take the reading D
 Difference (D - C) is called a staff intercept S

Horizontal distance

If the reduced level of A is xm above datum, the reduced level of –

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Stadia system

This is sometimes called the fixed hair Tacheometry because it involves the use of additional horizontal
lines or hairs marked on the diaphragm of the theodolite.

The two (2) stadia hairs or lines (upper and lower) are places equidistant, above and below the main
horizontal cross hair.

The system employs both a theodolite and staff and has one telescopic pointing with three simultaneous
reading taken (lower, middle, and upper)

The distance between the stadia hairs is fixed and is called stadia interval.

In the telescopes field of view, the stadia hairs subtend a certain length of the staff, S, called the staff
intercept.

The staff intercept(S) varies or is proportional to its distance from the instrument; and from the above

general principle, . ⁄ ( ⁄ )/ .

Basic principle of stadia system

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The principle of this form of tacheometry, in which the parallactic angle 2α remains, fixed and the staff
intercept S varies with distance D, is shown above. The parallactic angle is defined by the position of the
stadia hairs, c and e, each side of the main cross-hair b, then by similar triangles:

(⁄)

NB. In modern telescopes f and i are so arranged that c = 100.

Equation (1) above is basically correct for horizontal sights taken with any modern instrument.

For Detailed examination

In stadia tacheometry the line of sight of the tacheometer may be kept horizontal or inclined depending
upon the field conditions
Three cases considered for stadia system.
In the case of horizontal line of sight (Fig. 2.6), the horizontal distance between the instrument at A and
the staff at B is where
k and c = the multiplying and additive constants of the tacheometer, and
s = the staff intercept,
– , where ST and SB are the top hair and bottom hair readings, respectively
i.e.
1. Horizontal line of sight with a vertically held staff (normal)

is the focal length of the object lens System (telescope), d is the distance from the object lens to the
centre of the instrument, ce is the stadia interval i, and D is the distance from the staff to the centre of the
instrument; then by similar triangles:

. ⁄ / (⁄)

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 Engineering surveying II

But ( ) (⁄) ( )
The value ( f + d) is called the additive constant, K and ( f/i) is called the multiplying constant, c.

Alternatively

()

( ⁄ ) ( )

() ( ) ( ⁄)

( ⁄) ( ) (⁄)

(⁄) ( )

NB. - By adopting an anallatic lens in the telescope of the tacheometer, the multiplying conatant c is
made 100, and the additive constant zero(0).

- The value of k and c are kept equal to 100 and 0 (zero), respectively, for making the
computations simpler. Thus D = 100 s
- However in some tacheometers the additive constants are not zero, but vary from 30 cm to 60cm
actually specified by the manufacturers.
- S = The staff intercept, , where and are the top hair and bottom hair readings,
respectively.

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The elevations of the points,

These are determined by the height of instrument and taking the middle hair reading. Let
hi = the height of the instrument axis above the ground at A,
hA, hB = the elevations of A and B, and
SM = the middle hair reading
Then the height of instrument is

2. case 2: Inclined line of sight and staff held vertically

are readings given by the three lines and are those which would be given if the staff
were normal to the line of sight; ⁄ ( )

( ) ⁄

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Therefore

Horizontal distance

( )

Vertical distance

( )

The additive constant in modern instruments is reduced to zero, so the above formulae reduce to:

NB. When booking the vertical angle θ; the following convention is used.

- For Elevation i.e. sight uphill; θ is positive [+ve]

- Depression i.e. sight downhill; θ is negative [-ve]
- Vertical angle; is that angle between a given direction and the horizontal plane defined by E–W
axis

Reduced levels

With reference to the Figure above it can be seen that, given the reduced level of X (RLx);
Then the level of Y ( )
If the sight had been from Y to X then a simple sketch as in Figure 2.48 will serve to show that

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( )

NB.

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Measurements of tacheometric constants at the field

The multiplying constant (c = 100) and the additive constant (K = 0) are usually supplied with the
instrument by the manufacturer but they may vary due to ageing of the theodolite or temperature
variations.
If necessity arises to check them or determine them in an old tacheometer, the following method may be
employed.
 Set up the instrument on fairly level ground giving horizontal sights to a series of pegs at known
distances, D, from the instrument.
 Using the equation and substituting values for D and S, the equations may be
solved by;
i. Simultaneously in pairs and the mean taken.
ii. As a whole by the method of least squares.

For example: given

Measured distance (m) 30 60 90 120 150 (D-values)
Staff intercept (m) 0.301 0.600 0.899 1.202 1.501 (S-values)
Determine the value of C and k

From which C = 100 and k = 0 by either of the above methods.

Errors in the region of 1/1000 can occur in the constants.

Example 2
The following readings where taken with a theodolite onto a vertical staff
Stadia readings vertical angle horizontal distance
0.796 1.024 1.251 0o 45.736
1.873 2.179 2.485 5o00’ 61.013

Calculate the tacheometric constants

Solution

( ) ( ) ( )
()

( ) ( ) ( )
( )

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Accuracy and sources of errors

The accuracy depends on the instrumental and field errors

Instrumental errors

 The value of the instrument constants may not be correct.

- Due to the error in the construction of the diaphragm.
- Due to the assumption that modern telescopes are anallatic, when both c and k are
variables.

To meet the situation, the multiplying constant of the instrument should be found out by
careful field observation before commencement of the work.

 The graduations of the staff or stadia rod may not be uniform. To eliminate, the imperfect staff or
rod should be replaced, or necessary corrections should be applied.
 The adjustment of the tacheometer may not be perfect – permanent adjustments be checked

Field errors

These are gross errors, systematic errors and include:

 Incorrect centering and leveling. To avoid ensure perfect centering, and leveling of the plate
bubble and altitude bubble.
 Wrong staff readings are more common because of the number of readings required and the fact
that interpolation of the staff graduations have to be made each time resulting to errors in the staff
intercept, S.

NB. - The reading accuracy decreases as the horizontal distance D increases and the maximum
length of a tacheometric sight should be in the range of 100 – 120m.

- The surveyor should detect gross errors by comparing the difference between the top
and the middle reading, the differences should be the same otherwise the readings must
be repeated.
 Systematic errors; - Non perpendicularity of the staff is another serious source of error in D
- Differential refraction; Density of the atmosphere changes fairly rapidly closer to the earth
with a result that the lower line of sight is refracted more than the upper. To minimize this
effect, the lower staff reading should be less than 1.0 m.
- Reading the vertical circle (angle) of the theodolite with error.

Field work/procedure

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 Engineering surveying II

 A network of ground stations to be occupied by the instrument is set out on the ground
coordinated by traversing and their RLs obtained. It must be remembered that on unrestricted
sites, the maximum distance in tacheometric observations should be between 100 – 120m.
 Set the instrument over a station, center and have all necessary adjustments done.
NB: vertical and horizontal angles are measured on only one face. (Face left)
 Measure and record the height of the trunnion axis above the station mark (Hi).
 Select a suitable station as a reference object (RO), sight the point and record the horizontal
angle set for this direction. All the detail in the radiation pattern will be fixed in relation to the
direction of RO.
0
NB: The direction of the RO is set to 0 if details are to be plotted from radiation or the bearing
of the RO is set if details are to be plotted from coordinates.
 The staff man takes the first position of detail, and the telescope is rotated clockwise to bisect the
staff with the vertical hair in the field of view. Read and record the upper, middle and lower hair
readings.
NB: a check can be applied to the stadia readings since the center or middle reading should
be the mean of the lower and upper reading to with in .
 Still in this direction of the detail point, read and record the vertical and horizontal angles. The
angles should be taken to the accuracy of only.
 Signal the staff man to move to the new detail point.
 The procedure is repeated until when the observations of points near the instrument position is
completed.

NB: it is better to select each detail point in a clockwise direction in order to minimize the movement of
the staff man.

Check: the final sighting before the instrument is transferred should be to the RO to check that the setting
of the horizontal circle has been altered during observations. Should any alterations be
discovered, then all the horizontal circle readings are unreliable and should be re – measured. It
is therefore advisable that, during a long tacheometric observation a sighting to the RO be made
after say every 20 points of detail.

Tacheometric field book

Various methods can be used but a suitable method for most work is as shown in the table below.

All the information in the columns of bearings, V.A, U.H & L.M, M.H and remarks are recorded in the field.
The remainders are then computed in the office at latter stage.

Vertical circle readings are those read directly on the instrument, the reduction for is then done later in
the office to determine the angle of elevation or depression.

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NB: The vertical angle system of the instrument must therefore be understood at the beginning; either
it is vertical angle type or a zenith type.

A sketch is necessary during the booking. The sketch should include information such as type of
vegetation, widths of roads and railways, diameter of trees, heights and types of fences etc.

Plotting

 If radiation method is to be used then a network of control stations is first plotted on the
coordinate grid. The details and spot heights are then plotted using a protractor and a scale rule,
with the protractor oriented to the direction of RO. Each point of the detail is then plotted by
scaling an appropriate distance along each direction.
 If coordinates method is to be used then the details are plotted from their coordinates that were
computed.

The plotted points are shown on the plan by a circled dot against which is written the RL.

Using the field sketches, the details are filled in between these points and contours drawn by
interpolation.

The required details are then traced out leaving out unnecessary details such as construction or plotting
marks.

Examples

An instrument at A, sighted onto vertical staff held at B and C in turn, gave the following readings in the
table below

Sight Horizontal circle Vertical circle Staff readings (m)

B 0.442 0.744 1.045
C 0.655 0.960 1.265

If the instrument constants are C = 100, K = 0, and the height of instrument 1.52m; calculate the gradient
of the straight line BC.

Solution

( )

( )

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 Engineering surveying II

( )

( )

( )

( )

900
A

Example 2

The following readings were taken by a theodolite from station B on to stations A, C and D.

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 Engineering surveying II

Sight Horizontal circle Vertical circle Staff readings (m)

Top Centre Bottom
A
C 1.044 2.283 3.522
D 0.645 2.376 4.110

The line BA in Figure below has a bearing of 28°46’ and the instrument constants are 100 and 0. Find the
slope and bearing of line CD.

Solution

( )

( )

( )

( )

( )

( )

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 Engineering surveying II

√ ( )

( )

Bearing CD

( )

( ) ( )

Example 3

The following observations were taken with a tacheometer, having constants of 100 and zero, from a
point A to B and C. The distance BC was measured as 157 m. assuming the ground to be a plane within
the triangle ABC, calculate the volume of filling required to make the area level with the highest point,

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assuming the sides to be supported by vertical concrete walls. Height of instrument was 1.4 m, the staff
held vertically.

( )

( )

( )

( )

( )

( )

157 m
A

√ ( )( )( )

√ ( )( )( )

( )

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 Engineering surveying II

( )

Depth of fill at B = 31.44 m

Depth of fill at C =

( )

Example 4

The table shows tacheometric field observations involving the four corners of a plot, where in each case
the staff was held vertical with the instrument constants 100 and 0. Stations A and B were previously
coordinated as: A (152.48mN, 200.61mE) and B (107.55mN, 224.03mE),

Instrument Instrument Staff Bearing Vert. circle reading Staff readings

station height station
A 1.400 X 1.103, 1.403, 1.703
B 1.500 y 0.920, 1.142, 1.365
Calculate:

a) Area of the plot using the coordinate method

b) The uniform slope of line XY in the form 1: N

Solution

( ) ( )

( )

( )

√ √ ( )

( )

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 Engineering surveying II

( )

( )

( )

( )

( )

( )

( )

( )

( )

Example 5
The table shows tacheometric field observations that were interrupted by a sudden fall of a very big tree
along line QR, in which the surveyor had to re organize himself, technically in order to meet the intended
objective.

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 Engineering surveying II

Instrument Instrument Staff Bearing of Vert. circle reading Staff readings

station height station staff
P 1.550 Q 0.453, 1.564, 2.675
Q ? R ? ? ? ?
R 1.450 P 1.004, 2.503, 4.004
Calculate the uniform slope of line QR in the form of 1 in N, taking the instrument constants as 100 and 0
Solution

( )

( ) ( )

( )

( )

( ) ( )

( )

From sine rule;

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 Engineering surveying II

( )

( )

Reduced levels
Of Q;

( )
Of R is got from

( )

Slope of line QR

( )

Example 6
Readings were taken on a vertical staff held at points A, B and C with a tacheometer whose constants
were 100 and 0. If the horizontal distances from instrument to staff were respectively 45.9, 63.6 and 89.4
m, and the vertical angles likewise +5°, +6° and –5°, calculate the staff intercepts. If the mid-hair reading
was 2.100 m in each case, what was the difference in level between A, B and C?
(Answer: SA = 0.462, SB = 0.642, SC = 0.900, B is 2.670 m above A, C is 11.835 m below A)

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 Engineering surveying II

Solution

( )

( )

( )

( )

( )

( )

( )

( )
Example 4
A theodolite has a multiplying constant of 100 and an additive constant of zero. When set
1.35 m above station B, the following readings were obtained.
The coordinates of A are E 163.86, N 0.0, and those of B, E 163.86, N 118.41. Find the coordinates of C
and its height above datum if the level of B is 27.3 m AOD.
(Answer: E 2.64 N 0.0, 101.15 m AOD)

Earthworks

Introduction

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 Engineering surveying II

Earthworks refer to engineering works created through the moving of massive quantities of soil or uniform
rock.
Earthwork operations involve the determination of volumes of material that is to be excavated or
embanked in engineering project to bring the ground surface to a predetermined grade.
Estimation of areas and volumes is basic to most engineering schemes such as route alignment (roads,
railways), reservoirs, tunnels, dams, dikes, buildings, canals etc. The excavation and hauling of material
on such schemes is the most significant and costly aspect of the work, on which profit or loss may
depend.
Areas may be required in connection with the purchase or sale of land, with the subdivision of land or with
the grading of land.
Earthwork volumes must be estimated to enable route alignment to be located at such lines and levels
that cut and fill are balanced as far as practicable; and to enable contract estimates of time and cost to be
made for proposed work; and to form the basis of payment for work carried out.
Areas

In the context of surveying, the term area refers to a tract of land projected upon the horizontal plane
expressed in;

2
 Square metres (m ),
2 2
 Hectares (1ha = 10,000 m ; 100ha = 1m )
2
 Acres (1 acre = 4840 m )

Areas are involved in computation of volumes. Area calculations are to be considered under the following
headings;

1) Mechanical integrating device i.e. Planimeter

2) Areas enclosed by straight lines
3) Irregular figures

The planimeter

This is a mechanical device for determining the area of any plane shape (both regular & irregular)

The area is obtained from the measuring unit, consisting of an integrating disc which revolves and alters
the reading as the tracing point is moved around the perimeter of the figure.

2
On a fixed tracing arm instrument the readings are obtained directly in mm and converted according to
the plan scales to obtain the ground area.

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On a movable arm instrument, the tracing arm length can be set to particular values depending on the
plan scale so that the readings obtained give the ground area directly.

In the normal way, needle point F is fixed outside the area to be measured, the initial reading noted, the
tracing point traversed around the area and the final reading noted. The difference of the two readings
gives the number of revolutions of the measuring wheel, which is a direct measure of the area. If the area
is too large to enable the whole of its boundary to be traversed by the tracing point
P when the needle point F is outside the area, then the area may be sub-divided into smaller more
manageable areas, or the needle point can be transposed inside the area.
The area of plan is calculated from the following formula when using Amsler polar planimeter.

( )

Where M - The multiplying constant of the instrument; its value is marked on the tracing arm of the
instrument,
RF and IR - The final and initial readings,
N - The number of times the zero mark of the dial passes the index mark.
C - The constant of the instrument marked on the instrument arm just above the scale divisions. Its value
is taken as zero when the needle point is kept outside the area. For the needle point inside,

Example 1

What is the area of a piece of land which has a plan area of 1613mm2 as measured by a fixed – arm
planimeter if the scale of the plan is 1/2500.

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Example 2

Across sectional area was measured using a fixed arm planimeter which gave readings in mm2. The
initial planimeter reading was 620 and the final reading was 8004. Given that the horizontal scale of the
cross section was 1 in 200 and the vertical scale 1 in 50, calculate the true area represented by the cross
section.

( )

Areas enclosed by straight lines

The area of the tract of the land is computed from its plan which may be enclosed inside the survey lines
of a tape and offset survey or theodolite traverse and the fields enclosed by straight, irregular or
combination of straight and irregular boundaries. When the boundaries are straight the area is determined
by subdividing the plan into simple geometrical figures such as triangles, squares, circles, rectangles,
trapezoids, etc.

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 Engineering surveying II

Triangle Rhombus Ellipse

a
h h
b

b
b

Trapezium
Circle Sector
a

r
√ ( )( )( ) Ɵ h

( )

Example1
A page of the field book of a cross staff survey is given below. Plot the required figure and calculate the
relevant area.

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 Engineering surveying II

Solution

Identifying standard figures

1) ( )( )

2) ( ) ( ) ( )

3) ( ) ( ) ( )

4) ( )( )

5) ( ) ( ) ( )

6) ( ) ( ) ( )

7) ( ) ( ) ( )

8) ( ) ( ) ( )

Areas from coordinates

For a traverse survey, plotted from coordinates, the area enclosed is determined from its coordinates.

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 Engineering surveying II

Consider a closed traverse ABCA and the required area is ABC with
coordinates ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )

( )

*( ) ( )+

*( ) ( )+

* +

Example1
Determine the area in hectares enclosed by the line of a closed traverse survey ABCDE from the
following data.
Station E(m) N(m)
A 100.00 200.00
B 206.98 285.65
C 268.55 182.02
D 292.93 148.80
E 191.74 85.70

Solution

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 Engineering surveying II

Station N(m) E(m) N1E2(m2) (+ve) N2E1(m2)(-ve)

A 200.00 100.00 28565.00

B 285.65 206.98 41396.00 37674.50

C 182.02 268.55 76711.31 39960.24

D 148.80 292.93 53319.12 25104.10
E 85.70 191.74 28530.91 38348.00
A 200.00 100.00 8570.00
Σ 208527.34 169651.84

208527.30
less 169651.84
38875.46
2
Area ABCDEA = ½(ΣN1E2 - ΣE2N1) 19437.73 m

Area in hectares
19437.73
Area ABCDEA   1.944ha
10000
Example 2
The coordinates below refer to a closed traverse PQRS
Station E(m) N(m)
P +35.2 +46.1
Q +162.29 +151.0
R +14.9 +218.6
S -69.2 -25.2
Calculate the area in hectares enclosed by the stations
Solution

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 Engineering surveying II

2 2
Station N(m) E(m) N1E2(m ) (+ve) N2E1(m )(-ve)

P 46.10 35.20 5315.20

Q 151.00 162.90 7509.69 35609.94

R 218.60 14.90 2249.90 -375.48

S -25.20 -69.2 -15127.12 -3190.12
P 46.10 35.20 -887.04 35.20
Σ -6254.57 37394.74

-6254.57
-37394.74
-43649.31
2
Area ABCDEA = ½(ΣN1E2 - ΣE2N1) -21824.66 m

Area in hectares
21824.66
Area ABCDEA   2.18ha
10000
Alternatively

 46.1  162  151  14.9   218.6  69.2    25.20  3.2 

1
Area ABCDEA 
2

1
151  35.2  218.6  162.9   25.2  14.9  46.1  69.2
2
 21824.66m 2

Exercise
1) The coordinates below refer to a closed traverse ABCDEA
Station E(m) N(m)
P +51.0 -150.2
Q +300.1 -24.6
R 220.1 151.3
S -125.2 -51.1
Calculate the area in hectares enclosed by the stations (Ans. 9.3299 ha)

2) What area in acres is bounded by the traverse ABCDEFG whose metric coordinates are given
2
below (Ans. 628570.14m )

Station E(m) N(m)

A 98.3 422.2

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 Engineering surveying II

B 326.0 940.6
C 678.9 1024.4
D 914.9 1276.0
E 1388.5 812.5
F 860.4 306.7
G 492.3 535.8

Area Enclosed by Irregular Boundaries

For irregular boundaries, they are replaced by short straight boundaries, and the area is computed using
approximate methods or Planimeter when the boundaries are very irregular. Standard expressions as
given below are available for the areas of straight figures.
The Two fundamental rules for the determination of areas of irregular figures are
(i) Trapezoidal rule and
(ii) Simpson's rule.
Trapezoidal rule
Consider an area bounded by a survey line and boundary below

In trapezoidal rule, if the distance d is short enough for the length of the boundary, the boundaries
between pairs of the offsets is assumed to be straight. Therefore the area shall be divided into a series of
trapezoids.
( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )

( )

*( ) ( )+

*( ) ( )+

*( ) ( )+

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 Engineering surveying II

Simpson's Rule
In Simpson's rule it is assumed that the irregular boundary is made up of parabolic arcs. The areas of the
successive pairs of intercepts are added together to get the total area.
It is essential that the portion of area under consideration is divided into a number of even strips (i.e. odd
number of ordinates)

Consider strips (1) and (2) area contained between offsets O 1 and O2.

( )( ) ( )

( )
( ) ( )

( ) ( )

{( ) ( )}

{ } *( ) +

*( ) +

*( ) + *( ) +

*( ) +

*( ) +

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 Engineering surveying II

Comparison between Trapezoidal rule and Simpson’s rule

Similarities
 The intervals between the offsets or the strips are all equal
 Both are applied to determine areas and volumes of irregular boundaries
Differences between Simpson’s rule and Trapezoidal rule
Simpson’s rule Trapezoidal rule
 The boundary between the ordinates is  The boundary between the ordinates is
considered to be a parabolic arc. considered to be straight.
 Gives more accurate results  Gives approximate results
 Rule applies only to odd ordinates or even  Rule has no limitation. Applies to any
strips. number of ordinates

Example1
(a) The following perpendicular offsets were taken from a chain line to a hedge:
Distance(m) 0 6 12 18 24 30 36
Offset(m) 5.40 4.50 3.60 2.70 1.80 2.25 3.15

Calculate the area enclosed between the chain line and the offsets by
(i) Trapezoidal rule
(ii) Simpson’s rule
a) The following perpendicular offsets were taken from a chain line to a hedge:
Distance(m) 0 5 10 15 20 30 40 50 65 80
Offset(m) 3.40 4.25 2.60 3.70 2.90 1.80 3.20 4.50 3.70 2.80

Calculate the area in hectares by (i) Trapezoidal rule (ii) Simpson’s rule

Solution

Using *( ) ( )+

*( ) ( )+

( )

Using *( ) +

*( ) ( ) ( )+

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 Engineering surveying II

* +

Alternatively

Simpson's Rule Trapezoidal rule

Distance Offset F + L 4E 2O Distance Offset F+L 2 x Rest
0 5.40 5.40 0 5.40 5.40
6 4.50 4.50 6 4.50 4.50
12 3.60 3.60 12 3.60 3.60
18 2.70 2.70 18 2.70 2.70
24 1.80 1.80 24 1.80 1.80
30 2.25 2.25 30 2.25 2.25
36 3.15 3.15 36 3.15 3.15
F+L 8.55 9.45 5.40 F+L 8.55 14.85
4E 37.80 x 4.0 x 2.0 2Rest 29.70 x 2.0
2O 10.80 37.80 10.80 38.25 29.70
57.15 x 3.0
d
x 2.0 A  F  L  2Rest) 114.75
d
A  F  L  4E  20 114.30 2
3

(b) Using *( ) ( )+

*( ) ( )+ *( ) ( )+

*( ) ( )+

( ) ( ) ( )

(c) Using *( ) +

*( ) ( ) ( )+ *( ) ( )+

*( )+ *( ) ( )+

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 Engineering surveying II

Distance Offset F+L 4E 2O Distance Offset F+L 2 x Rest

0 3.40 5.40 0 3.40 3.40
5 4.25 4.25 5 4.25 4.25
10 2.60 2.60 10 2.60 2.60
15 3.70 3.70 15 3.70 3.70
20 2.90 2.90 20 2.90 2.90
F+L 8.30 7.95 2.60 F+L 6.30 10.55
4E 31.80 x 4.0 x 2.0 2Rest 21.10 x 2.0
2O 5.20 31.80 5.20 27.40 21.10
45.30 x 2.5
d
x 1.67 A1  F  L  2Rest)
 68.50
d 2
A  F  L  4E  20 75.50
3
20 2.90 2.90
20 2.90 2.90 30 1.80 1.80
30 1.80 1.80 40 3.20 3.20
40 3.20 3.20 50 4.50 4.50
F+L 6.10 1.80 0.00 F+L 7.40 5.00
4E 7.20 x 4.0 x 2.0 2Rest 10.00 x 2.0
2O 0.00 7.20 0.00 17.40 10.00
13.30 x 5.0
d
d
x 3.33 A1  F L  2Rest)
 87.00
A  F  L  4E  20 2
3 44.333
50 4.50 4.50
50 4.50 4.50 65 3.70 3.70
65 3.70 3.70 80 2.802.80
80 2.80 2.80 F+L 7.30 3.70
F+L 7.30 3.70 0.00 2Rest 7.40 x 2.0
4E 14.80 x 4.0 x 2.0 14.70 7.40
2O 0.00 14.80 0.00 d
x 7.5
22.10 A1  F  L  2Rest)
 110.25
2
d
x 5.00
A F  L  4E  20
3 110.5
Using Trapezoidal Rule

Using Simpson's rule A  68.50  87.00  110.25  265.75m 2

 3.20  4.50 
A  75.50  44.33  110.50    10  268.83m
2

 2 

Cross sectional areas

Finding the areas of cross-sections is the first step in obtaining the volume of earthwork (cut or fill ) to be
handled in route alignment projects (road or railway), or reservoir construction.

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 Engineering surveying II

This involves calculating the area between the existing and the proposed formation level, at each cross
section drawn.

Case 1: level section

b = Finished road width at road or formation level

H = Centre height
W1, W2 = Side widths, measured horizontally from the centre-line and depicting the limits of the
construction
1 in S = Side slope of 1 vertical to S horizontal
1 in G = Existing ground slope

Example

The survey of a proposed cutting shows that the depths at 20m intervals are 0.0m, 0.9, 1.5, 3.2, and
0
3.3m. Given that the roadway is to be 5m wide and that the cutting has 45 side slopes; calculate

The plan surface area of the excavation

Solution

( ) ( ) ( )

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 Engineering surveying II

Chainage Depth(h) Top width F +L 4E 2O

5.0 +2h
0.00 0.00 5.00 5.00
20.00 0.90 6.80 6.80
40.00 1.50 8.00 8.00
60.00 3.20 11.40 11.40
80.00 3.30 11.60 11.60
F+L 16.60 18.20 8.00
4E 72.80 x 4.0 x 2.0
2O 16.00 72.80 16.00
105.40
A
20
F  L  4E  20 880
3 702.6667

Case 2: Level two

Existing Ground surface slopping

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 Engineering surveying II

Here the figure is split into regular shapes/triangles whose areas can be determined easily.

( )

( )

( ) ( )

Depth calculation

( )

( )

Plan width calculation

⁄ ⁄

( ) ⁄ ⁄ ( )

( )

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 Engineering surveying II

( )

( ) ( ) . /

⁄ ⁄

( ) ⁄ ⁄ ( )

( )

( )

( ) ( ) . /

The rule therefore is:

 When the two grades are running in opposing directions (as in AA’C), add (signs
opposite+−).
 When the two grades are running in the same direction (as in BB’D), subtract (signs
same).

Example

A roadway is to be built on ground having a traverse ground slope of 1 in 8. The road is 8.0m wide, has a
central height of 3.5m and 1 in 4 side slopes. Calculate the cross sectional area of the embankment.

Solution

Depths of fills

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 Engineering surveying II

( ) ( )

( ) ( )

Plan widths

( ) ( )

( )

( ) ( ) . /

And

( )

( )

Cross section areas

( ) ( ) {. /( ) }

( ) ( ) ( )

Level 2: Cut and Fill

This occurs when the road is being built around the side Of a sloping hill and is used for economic
reasons since the cut sections can be used in the fill section.

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 Engineering surveying II

Considering a cut section

( ) ( )

( )

. /( ) . /

Plan/side width

( ) ( ) . /

Area of the cut

. /

. / . / {. / . /}

{ . / }

. /
{ }

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 Engineering surveying II

Considering the fill section

Plan / side width

()

( ) ( )

() ( )

( )

Therefore fill depth

. /

Plan /side width

. / ()

Area of filling

. /

. / . / {. / . /}

{ . / }

. /
{ }

Example 1

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 Engineering surveying II

Calculate the side widths and cross-sectional areas of cut and fill on a hillside section shown below,
having the following dimensions:

Road width = 20m existing ground slope = 1 in 5 (20%)

Side slope in cut = 1 in 1 (100%) centre height in cut = 1m
Side slope in fill = 1 in 2 (50%)

Solution

Side widths

( ) ( ) . /

( ) ( )

. /

( )

Cross section areas

. /
{ }

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 Engineering surveying II

. /
{ }

. /
{ }

. /
{ }

Example 2

Calculate the side widths and cross-sectional areas of cut and fill on a hillside section shown below,
having the following dimensions:

Road width = 15m existing ground slope = 1 in 5

Side slope in cut = 1 in 1 centre height in fill = 0.5m
Side slope in fill = 1 in 2

Solution

Side widths

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 Engineering surveying II

( ) ( ) . /

( ) ( )

. /

( )

Cross section areas

. /
{ }

. /
{ }

. /
{ }

. /
{ }

Volume Calculations

Volume is the quantity or space occupied by earthwork in question. The importance of volume
assessment has already been outlined. Many volumes encountered in civil engineering appear, at first
glance, to be rather complex in shape. Generally speaking, however, they can be divided into prisms,
wedges or pyramids.
(1) Prism
The two ends of the prism are equal and parallel, the resulting sides thus being Parallelograms.

In this case, from prismoidal formula

( )

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 Engineering surveying II

A1 = Am = A2:

( )

(2) Wedge

( )

,( ) -

In this case Am is the mean of A1 and A2, but . Thus ⁄ :

( )

(3) Pyramid

Volume of pyramid

( )

NB: (i) A prismoid is a solid figure with plane parallel ends and plane sides.

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 Engineering surveying II

(ii) Thus, any solid which is a combination of the above three forms and having a common value for

L, may be solved using equation ( ) . Such a volume is called a prismoidal and

the formula is called the prismoidal equation

(iii) The prismoidal equation is correct when the figure is a true prismoid. In practice it is applied by
taking three successive cross-sections. If the mid-section is different from that of a true prismoid, then
errors will arise.
(iii) Prismoidal formula is comparable to Simpson’s rule for a number a series of cross sections of a

constant distance a part. ( )

Examples

1. A new road is to be constructed with formation width of 20 m, side slopes of 1 vertical to 2.5
horizontal. The heights of fill at the center line of three successive cross-sections, 50 m apart, are
3.3 m, 4.1 m, and 4.9 m, respectively. The existing ground has a cross fall of 1 in 10. Calculate
the volume of the fill.

2. Using prismoidal formula ( ( )), calculate the volume of material to be removed

for a road length of 30m; Formation width 8.0m, depth of commencement 8.0m; depth at the end
5.0m, side slopes 1 in 1
Solutions

End-area method
If only two cross sectional areas A1 and A2 are a horizontal distance L apart, the volume contained
between them is given by;

End area method is comparable to the trapezoidal rule; for a series of cross sections, with a horizontal
distance d a part, the total volume v is expressed as below

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 Engineering surveying II

( ) ( ) ( ) ( )

( )

*( ) ( )+

Comparison of end-area and prismoidal equations

Consider the Figure above, then

In order to compare the methods, the volume of Figure, will be computed as follows:
Dimensions are;
Centre heights: h1 = 10m, h2 = 20m, hm = 18m
Road widths: b1 = b2 = bm = 20m
Side slopes: 1 in 2 (50%)
Horizontal distance between sections: l = 30m, L = 60m
N.B: For a true prismoid hm would have been the mean of h1 and h2, equal to 15 m. The broken line
indicates the true prismoid; the excess area of the mid-section is shown tinted.
Solution
Top width
Depth (h)
10 60
80

18 92
20 100

(1) The true volume is thus a true prismoid plus two wedges, as follows:

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 Engineering surveying II

* ( ) +

,( ) - ,( ) -

,( ) - ,( ) -

(ii) Volume by Prismoidal rule (Am will now have centre height h = 18m)

* ( ) +

This error is approximately equal to the area of the excess mid-section multiplied by , i.e.
( ) , and is so for all such circumstances; it would be negative if the mid-area had been
smaller.
(iii) Volume by end area

Thus, in this case the end-area method gives a better result than the prismoidal equation. However, if we
consider only the true prismoid, the volume by end areas is 46 500m3 compared with the volume by
3
prismoidal equation of 46 000m , which, in this case, is the true volume.
Therefore, in practice, it can be seen that neither of these two methods is satisfactory. Unless the ideal
geometric conditions exist, which is rare, both methods will give errors. To achieve greater accuracy, the
cross-sections should be located in the field, with due regard to the formula to be used.

Volume from spot heights

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 Engineering surveying II

CURVE RANGING

Introduction

A curve is a line or an outline that gradually deviates from being straight for some or all of its length; i.e.
it’s a parabolic line

Ranging is the process of defining a straight line by use of a series of ranging rods.

Curves are set out for different purposes, like alignment of a project involving roads or railways, a kerb
line at a junction or a shape of an ornamental rose bed in a town centre.

They are required to connect two or more straights so that vehicles can pass smoothly from one straight
to another.

So generally the curves are used where the direction of a line may change due to some unavoidable
circumstances.

The angle of the change in direction is known as the deflection angle

Classification of curves

 Horizontal curves
 Vertical curves
Horizontal Curves;

These are curves whose measurements, design and construction are all considered in the horizontal
plan.

Types of horizontal curves

 Circular curves
 Transition curves
Circular Curves

These are curves of constant radius.

Sub division of circular curves

o Simple circular curves

These are curves consisting of a single arc of constant radius connecting the two straights.

o Compound circular curve

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 Engineering surveying II

This is a curve consisting of two or more circular arcs of different radii which lie on the same side of a
common tangent.

The centers of the different arcs lie on to the same side of their respective tangents

o A reversed circular curve

This is a circular curve consisting of two or more arcs bending in opposite directions. Their centers lie on
opposite sides of the curve or common tangent. Their radii may either be equal or different.

TERMINOLOGIES OF CIRCULAR CURVES

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 Engineering surveying II

TP1 – TP2 is the circular curve.

R - The radius of curvature

TP1 and TP2 - Are tangent points

IP - The intersection point∙
is the deflection point
TP1 - IP and IP – TP2 - is the tangent length * +
a - is the external distance∙
b - is the mid ordinate∙
<TP1 IP TP2 is the intersection angle

Preliminary setting out information

( )

⁄ ( )

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 Engineering surveying II

( )

Relationships in circular curves

i. The tangential angle α at T to any point, x, on the curve TU is equal to half the angle subtended
at the center of curvature O, by the chord from T to that point∙
Similarly;

ii. The tangential angle β at any point on the curve is equal to any forward point, Y, on the curve is
equal to half the angle subtended at the center at the center by the chord between the two points
X and Y ∙
Combining ( ) and ( )

, - , and it follows that ( ), thus:

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 Engineering surveying II

iii. The tangential angle to any point on the curve is equal to the sum of the tangential angles from
each chord up to that point.
The above relationship illustrated in above diag. is used in setting out the curves by the methods of
tangential angles∙

Specification of a curve

Circular curves can be specified in 3 ways

 By Versed sine – length from the centre of the chord to circumference

 As Radius Curve i.e. in terms of their radius, for e.g. a 600m radius curve
 Degree Curve i.e. in terms of the angle subtended at the center of curvature by a 100m arc, e.g. a
o
2 curve.
o o
XY 100m and subtends an angle of D at the center O. The curve TU is therefore a D curve.

NB: The smaller the value of D, the less shaper the curve and vice versa.

Relationship between the Radius and Degree Curves

( )

Length of Circular Curve ( )

For a radius curve, ( ) where R is in meters and in radians

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 Engineering surveying II

For a degree curve, ( ) where and D are in the same units, i.e. degrees or radians.

Two straights AX and XB are to be connected by a 400m radius circular curve. The bearings and lengths
of the straights are as follows:

Straight WCB Length/m

Calculate:

a) The deflection or intersection angle

b) The chainages of the tangent points
c) The initial and final sub chords if 20m chain length is to be used.
d) The number of standard chord lengths
Solution

a) The deflection Angle

b) Chainage of Tangent points

But Tangent length . /

()

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 Engineering surveying II

( )

c) Calculation of Initial and Final sub chords

( )

()

( )

( )

d)

Example 2

Two straights BA and CA intersect at a point A which falls in the bed of a river. These are to be connected
by a simple circular curve of radius 200m. To do this, a line MN connecting these two straights at points M
and N respectively is measured to be 170m. The clockwise angles and .
The chainage of point,

Determine the;

(i) Chainage of tangent points

(ii) Length of the curve
Using 30m multiple chord lengths, prepare setting out data for curve by tangential method.

Solution

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 Engineering surveying II

In MA;

( )
Angle of deflection
Let and be the tangent points

Tangent length . /

Applying sine rule to

(i) Chainage of Tangent points

(ii)

a) Calculation of Initial and Final sub chords

( )

()

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 Engineering surveying II

( )

( )

b)

Chord Cummul. Tang.

S/no. Chainage length (c) Curve length angle

T 1607.39 0 0 0 0
1 1620.00 12.61 12.61 1.806249449 1.806249449
2 1650.00 30.00 42.61 4.297183463 6.103432913
3 1680.00 30.00 72.61 4.297183463 10.40061638
4 1710.00 30.00 102.61 4.297183463 14.69779984
5 1740.00 30.00 132.61 4.297183463 18.9949833
6 1770.00 30.00 162.61 4.297183463 23.29216677
7 1800.00 30.00 192.61 4.297183463 27.58935023
8 1830.00 30.00 222.61 4.297183463 31.88653369
9 1860.00 30.00 252.61 4.297183463 36.18371716
10 1890.00 30.00 282.61 4.297183463 40.48090062
11 1920.00 30.00 312.61 4.297183463 44.77808408
12 1950.00 30.00 342.61 4.297183463 49.07526755
13 1980.00 30.00 372.61 4.297183463 53.37245101
14 2010.00 30.00 402.61 4.297183463 57.66963447
15 2026.27 16.27 418.88 2.330505832 60.00014031

Example 3

The straights AI and IB deviate to the left by They are joined by a circular curve such that the
shortest distance between the curve and intersection point is 35.3m

Calculate

(i) The radius of the curve

(ii) Length of the long chord and major offset

Setting out circular Curves

This can be done using the following equipment

 Chain and Tape only – usually employed by labour based methods

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 One theodolite and Tape or chain

 Two theodolites
Methods of setting out curves

Small radius curves

This can be set quickly and accurately by using tapes or chains only employing the following methods

 Offsets from tangents

 Offsets from long chord
 Quarter method
Large radius curves

These can be set out by

 Tangential method – one theodolite and tape only

 Offsets from chords produced
 Two theodolites
Through Chainage

Through chainage is the distance usually in meters, measured from the starting point of the scheme to a
particular point in question. It is used in road, railway, pipe line and tunnel construction as a means of
reference to any point on the center line.

Z is known as the position of zero chainage. Chainage continues to increase from Z along the center line.
At T chainage can continue to increase in two directions, either along the curve or along the straight to
point I.

If the chainage of the intersection point I, is known, the chainages of tangent points T and U, which both
lie on the center line can be found as follows

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Location of I, T and U in the Field

 Locate the two tangent lines AC and BD.

 Set up a theodolite on one line say AC, and sight towards the intersection of the two tangents at I.
 Set two pegs X and yon line such that BD will intersect line XY∙
 Join pegs x and y by means of a straight line∙
 Set up theodolite on line BD, pointing if towards the direction of D and fix I at appoint where the
line of sight bisects the string x y.
 Set up the theodolite over I and measure angle AIB, hence

 Calculate the tangent lengths IT and I U using

 Measure back from I to T and U a distance of

Location of T and U When I is Inaccessible

Procedures

1. Choosing points A and B somewhere on the tangent, such that it is possible to sight one another.
2. Measure distance AB.
3. Measure angles
4. Use the sine rule to calculate IA and IB.

5. Calculate IT and IU from

6. AT = IA IT and BU =IB –IU, hence set out T and U.

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If A and B are chosen to be on the other side of T and U, AT and BU will have negative values,
and measured backwards.

Offsets from chords produced

This method is used when the land between the two tangent points T & U need not to be accessible.

Assumptions

Procedures

 Arrange the chain along TI with its end at A1 is swung through a calculated offset A1A with T as
the centre to locate point A on the curve
 Pull the chain along TA produced until its ends are at A and B1 with A as the centre, swing the
end AB1, through a calculated offset B1B to locate point B on the curve.
 Repeat the process for other points
 As a check, the last point should coincide with the last tangent point U on the curve
 Calculate the offsets

Chord length

Also

( )

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B2 B is the offset from the tangent AB2 as for AA1 which is the offset from TI, we may derive B2B to be;

( )

( )

( ) ( )

( )

Example

A 300m radius curve is to be set out by offsets from chords produced. The chainages of the first and
second tangent points T = 327.5m and U = 425.3m respectively. Calculate the lengths of the offsets.
Setout pegs at even chainages of 20m

Solution

TANGENTIAL METHOD OF SETTING OUT CURVES

This is also termed as deflection angle method. The method employs one theodolite and tape or chain.

Assumptions

Procedures

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 Determine the total length of the curve ( )

 Set out tangential line
 Set up the theodolite at and with a telescope horizontal scale reading of 0 point the telescope
to I.
 Turn the telescope until you get the horizontal reading to give a point B on the curve and
chain distance from to B
 Turn the telescope until the horizontal reading is and chain distance from B to give C
on the curve.
 Repeat the above procedure but with the scale reading of
 As a check the tangential angle ITU should be equal to /2.

The angle subtended at the centre of of a circle is equal to twice that at the circumference

Therefore; chord

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Example;

It is required to connect two straights whose deflection angle is 13 16’ by a circular curve of radius
6000m. Make the necessary calculation for setting out the curve by tangential angles method if the
through chainage of the intersection point is 2745.72m. Use a chord length of 25m and sub chord at the
beginning and end of the curve to ensure that pegs are placed at exact 25m multiplies of through
chainage.

Solution

But Tangent length . /

(iii)

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c) Calculation of Initial and Final sub chords

( )

()

( )

( )

d)

Note

As a check the final cumulative tangential angle should equal to /2, in this example there is only 5
difference which is acceptable. Also the sum of the chords should equal the total length of the circular arc,
and in this example, there is no deference.

Since is a proportional to the chord length, chords of equal length will have the same tangential angle.

Two theodolites are used, one being at each tangent point ( ) The disadvantage of the two method
is that two of every thing is needed, e.g. two engineers, two instruments and two assistants to locate
pegs.

In the above fig point Z, is set out from T relative to TI and ( ) is set5 out from U relative
to UI the two line of sight will intersect at Z and a peg is driven in to the ground by an assistant.

For large curve, two way radios are very useful.

Transition Curves
A transition curve is a curve in which the radius changes continuously along its length and is used for the
purpose of connecting a straight with a circular curve.
The transition curve is used to achieve the following:
i. To reduce the centrifugal effect
ii. To reduce the tendency of vehicular skidding;
iii. To minimize passenger discomfort;
iv. To provide convenient sections over which super-elevation or pavement widening may be
applied;

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v. To improve the appearance of the road by avoiding sharp discontinuities in alignment at the end
and beginning of circular curves.
Types of transition curves

Composite transition curve

This is one in which the road section has an entrance and exit transition curve of two equal lengths
connecting a central circular arc of constant radius R.

The type of design has widespread use but its disadvantageous in that the radial force is constant and
maximum on the circular curve. If this maximum force is large, then the section represents a danger
length. The limiting radii for safety are therefore always stipulated by the depart men t of transport in the
design standards.

Wholly transition curves

Here, the road consists of only two transition curves of equal length connected at only one point (T c) of
maximum radial force i.e. when r = R, and hence increasing safety of the road.

Properties of Transitions

Where;

φmax = Spiral angle

θ = Deflection angle

T = Tangent length

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R = Radius of circular curve

S = Shift

L = Length of Spiral (or transition curve)

I = Point of intersection

T = Beginning of spiral

T1 = Beginning of circular curve

T2 = End of circular curve

U = End of spiral

Important properties of a spiral are:

( ) ( )

Length of Transition curve

The length of the transition should be determined from the following two considerations:

i) The rate of change of change of super-elevation should also be such as not to cause higher
gradients and unsightly appearances.
ii) The rate of change of centrifugal acceleration adopted in the design should not cause
discomfort to the drivers. If C is the rate of change of acceleration then:

( )

Where;

aT1 = Radial acceleration at T1. ⁄ /

aT = radial acceleration at T (= 0)

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t = time taken ( ⁄ )

Substituting the above in eqn. 1 gives

. ⁄ /
( ⁄ )

From which the length of transition curve, L is given by

and

()

3
The value of C is usually taken as 0.3m/s . But normally vary between (0.2 - 0.6)

Example
A two-lane (7.0m wide) pavement on a National highway in hilly terrain has a curve radius of 250m. The
o
design speed is 80kph, and the angle of deviation is 60 and the rate of change of radial acceleration is
3
0.3m/s . Determine the following
i) The length of transition curve;
ii) The tangent length
iii) The total length of the curve.

Super elevation

Inward tilt or transverse inclination given to the cross section of a roadway throughout the length of a
horizontal curve to reduce the effects of centrifugal force on a moving vehicle; expressed as a
percentage.

Maximum super-elevation Value, emax

It is desirable that the super-elevation should be such that a moderate amount of friction is developed
while negotiating flat curves and friction not exceeding the maximum allowable value should be
developed at sharp curves. Therefore designing the super-elevations to fully counteract the centrifugal
force developed at a fraction of the design speed will provide the necessary balance.
In Uganda the value is limited to 0.08 (8%) and for UK 0.07(7%)

Vertical curves
A vertical curve provides a smooth transition between successive gradients in the road profile.

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It is composed of a series of straight-line gradients connected by curves, normally parabolic along the
road profile.

Types of vertical curves

Crest/summit curves
When the algebraic difference in gradients, A is positive the curve is called a crest or summit curve.
Sag/trough curves
If A is negative the curve is called a sag or valley curve

Major Requirements of Vertical Curves

The two main requirements in the design and construction of vertical curves are the provision of:
 Adequate visibility, and
 Passenger comfort and safety.
Adequate visibility
This requirement is achieved by use of sight distances and K-Values. K –Value is the Ratio of the
minimum length of vertical crest curve in meters to the algebraic difference in percentage gradients
adjoining the curve.
Passenger discomfort

The effect of the radial force on the vehicle traversing a vertical curve must be minimized.

This can be minimized by:

 Restricting the gradients; this has the effect of reducing the radial force;
 Choosing a suitable type and length of curve such that this reduced force is introduced gradually
and uniformly as possible.

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Gradients
The rate of rise or fall of road surface along its length with respect to horizontal distance is termed as
gradient.
Gradients of up to about 7% have little effect on the speeds, road user costs through delays, extra fuel
costs and accidents of passenger cars.

The Uganda Road Design manual (2004), suggests maximum gradients as presented in Table below

Length of a vertical curve

The minimum length of vertical curve Lmin for any given road is obtained from the formula.

Where;
K = constant obtained from MoWH&C standards (K = R/100)
R = radius of curvature of the curve (in meters)
A = algebraic difference in grade (%)
Minimum Radii for Crest Curves as Recommended by MoWH&C

Minimum Radii for Sag Curves as recommended by MoWH&C

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Equation of a Vertical Curve

A simple parabola is recommended when modeling vertical curves. The parabola provides a constant rate
of change of curvature, and hence visibility, along its length. The vertical curve is of the form:

( )

( )
()

Where; ( )

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Vertical Curve Examples

Question one
The elevation of an intersection of rising gradient of 1.5% and a falling gradient of 1.0% on a proposed
road is 93.600m AOD. Given that the K-Value for this particular road is 55, the through stationing of the
intersection point is 0 + 671.340 and the vertical curve is to have equal tangent length. Calculate:
a) The through stationing of the tangent points of the vertical curve if the minimum required length is to be
used.
b) The elevations of the tangent points and the elevations at exact 20m multiples of through stationing
along the curve.
c) The position and level of the highest point on the curve.
Solution

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TOTAL STATION
Introduction
Basics of Surveying

 Surveying is the most important, oldest practice carried out, which includes, taking of
measurements on, below or above the surface of the ground and to plan the course of further
actions, based on the findings of the survey.
 The linear measurements and angular measurements are measured with the aid of few
instruments, viz., tape, chain, staff, dumpy level, theodolite, etc
 All the readings are noted in field books or survey records and kept as a hard copy and the
calculations are done later in office
 With such surveying instruments, survey work will be slow and tedious.

Invention of Modern Surveying

 As an outcome of continuous technological development, in the last few decades, new varieties
of electronic instruments have been invented.
 With these inventions, modern surveying practices came into existence.
 Hence with modern surveying instruments, survey work will be precise, faster and less tedious.

Among the newest computer-driven technologies available to assist surveying engineers in their work of
developing and processing spatial data are;

(1) Total station instruments, including robotic systems;

(2) The global positioning systems (GPS);

(3) Digital photogrammetry and light detection ranging (LIDAR);

(4) Satellite remote sensing; and

(5) Geographic information system (GIS).

These new systems are enabling surveying engineers to provide new and better types of information at
lower cost and in fraction of time previously required.

 All the readings are recorded automatically by the electronic instruments in a microprocessor and
based on these readings; critical values are calculated by the microprocessor, at real time.

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 Hence, the data collected in microprocessor is directly transferred and stored in the computer as
a softcopy.
 Hence modern surveying instruments are becoming more popular and they are gradually
replacing old surveying instruments such as compass and Dumpy level.

Modern Surveying Instruments

EDM-Electronic distance measurement

Total Station

The total station is an improvised version of modern surveying instruments such as EDM – Electronic
distance measurement, auto level and digital level.
Total station is a combination of an electronic theodolite and an electronic distance meter (EMD) and
software running on an external computer known as a data collector.

This combination makes it possible to determine the coordinates of reflector by aligning the instrument’s
cross hair on the reflector and simultaneously measuring the vertical and horizontal angles and slope
distances.
On board micro-processor in the instrument, takes care of recording, readings and the necessary
computations. The data can be easily transferred to a computer where it can be used to generate map.

Most modern total station instruments measure angles by means of electro-optical scanning of extremely
precise digital bar-codes etched on rotating glass cylinders or discs within the instrument.

The best quality total stations are capable of measuring angles down to 0.5 arc-second. Inexpensive &
quot; construction grade & quot; total stations can generally measure angles to 5 or 10 arc-seconds.

Measurement of distance is accomplished with a modulated microwave or infrared carrier signal,

generated by a small solid-state emitter within the instrument's optical path, and bounced off of the object
to be measured.

The modulation pattern in the returning signal is read and interpreted by the onboard computer in the total
station.

The distance is determined by emitting and receiving multiple frequencies, and determining the integer
number of wavelengths to the target for each frequency.

The typical Total Station EDM can measure distances accurate to about 3 millimeters or 1/1000th of a
foot.

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Robotic Total Station

The improvised total station by name ROBOTIC TOTAL STATION allows the operator to control the
instrument from a distance via remote control.
This eliminates the need of an assistant staff member, as the operator holds the reflector and controls the
total station from the observed point.

Components of a Total Station

o EDM
o Electronic theodolite
o On-Board Micro-processor
o Data Collector (built in or separate unit)
o Data Storage (internal or memory card)
o Prisms
 Micro-processor
o Averages multiple angle measurements
o Averages multiple distance measurements
o Computes horizontal and vertical distances
o Corrections for temp, pressure and humidity
o Computes inverses, polars, resections
o Computes X, Y and Z coordinates
 Specifications
o Range
 Reflector less –> 3 – 70 meters
 Single Prism -> 1 – 2000 m
 Triple Prism -> 1 – 2200 m

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o Accuracy
 Angles –> 1 - 5”
 Distance –> 3mm + 2ppm (prism)
 -> 4mm + 3ppm (reflector less)
o Data Storage
 2000 – 4000 points

Functions of Total Station:

 Coordinates determination:
Total station determines the coordinates of an unknown point relative to the known coordinate by
establishing a direct line of sight between the two points. Angles and distances are measured from the
total station to points under survey and the coordinates of surveyed points relative to the total station
position are calculated using trigonometry and triangulation. Some total stations have Global Navigation
Satellite System (GNSS), which does not require direct line of sight to determine coordinates.
 Distance measurement:
A total station has a small solid state emitter within the instrument’s optical path. They generate
modulated microwave or infrared signals that are reflected by a prism reflector or the object under survey.
The modulation pattern in the returning signal is read and interpreted by the computer in the total station.
The distance is thus determined by emitting and receiving multiple frequencies and determining the
integer number of wavelength, to the target, for each frequency.
 Angular measurement:
Most of the modern total stations have digital bar-codes on rotating glass cylinder that are installed within
the instrument. Angle measurements are done through electro-optical scanning of these digital bar-codes.
 Data processing:
The data recorded by the instrument may be downloaded from the theodolite to a computer and the
application software in turn generates a map of the survey area. Many advanced models of total station
have built-in micro-processor to record and compute distances, horizontal and vertical angles.

Advantages of Total Station

o Most accurate and user friendly.
o Gives position of a point (x, y and z) w. r. t. known point (base point).
o EDM is fitted inside the telescope.
o Digital display.
o On board memory to store data and compatibility with computers.
o Measures distance and angles and displays coordinates,
o Auto level compensator is available.
o Can work in lesser visibility also.
o Can measure distances even without prismatic target for lesser distances.

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o It is water proof.
o On board software are available.
o Total solution for surveying work.

Applications of total station in varieties of fields:

 Mainly used by land surveyors.
o Used by archaeologists to record excavations.
o By Police, crime scene investigators, private accident re-constructionists and insurance
companies to take measurements of scenes. . Once they take accurate measurements
with a total station they can use software to recreate the accident in a 3D animation
 In Civil Engineering Field Used For:
o General purpose angle measurements.
o General purpose distance measurement.
o Provision of control surveys.
o Contour and detail mapping.
o Setting out and construction work.
o Underground mining; a total station will be used to record the absolute location of the tunnel
walls ( stope ), ceilings (backs), and floors. This data can then be loaded into a CAD
program, and compared to the designed layout of the tunnel.

Auxiliary Equipment Required:

o Targets or Prisms to accurately define the target point of a direction measurement.
o A data recorder if one is not integrated into the total station.
o A download cable and software on a PC to capture and process the captured digital data to
produce contour and detail maps.
Conclusion
Total station makes the measurements more accurate.
Total station records the data on its inbuilt data recorder and generates maps within a fraction of time.
Hence time consuming will be less and Total station made survey work easy.
Hence TOTAL STATION IS A TOTAL SOLUTION FOR SURVEY WORK.
SATELLITE POSITIONING
Introduction
The concept of satellite position fixing commenced with the launch of the first Sputnik satellite by the
USSR in October 1957. This was rapidly followed by the development of the Navy Navigation Satellite
System (NNSS) by the US navy. This system, commonly referred to as the Transit system, was to provide
world-wide navigation capability for the US Polaris submarine fleet. The Transit system was made

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available for civilian use in 1967. However, as it required very long observation periods and had a rather
low accuracy, its application was limited to geodetic and navigation uses.
In 1973, the US Department of Defense commenced the development of NAVSTAR (Navigation
System with Time and Ranging) Global positioning system (GPS) and the first satellites were launched
in 1978. These satellites were essentially experimental, with the operational system scheduled for 1987.
Now that GPS is fully operational, relative positioning to several millimetres, with extremely short
observation periods of a few minutes, has been achieved. For distances in excess of 5 km GPS has been
shown to be more accurate than EDM traversing. It therefore has a wide application in engineering
surveying, with an effect even greater than the advent of EDM.
Apart from the high accuracies attainable, GPS offers the following significant advantages:
1) Position is determined directly in an X, Y, Z coordinate system.
2) Intervisibility between ground stations is unnecessary.
3) As each point is fixed discretely, there is no error propagation as in networks.
4) Survey points may therefore be selected according to their required function, rather than to
produce a well-conditioned network configuration.
5) Low skill required by the operator.
6) Position may be fixed on land, at sea or in the air. This latter facility may have a profound effect in
aerial photogrammetry.
7) Measurement may be carried out, day or night, anywhere in the world, at any time and in any
type of weather.
8) Continuous measurement may be carried out, resulting in greatly improved deformation
monitoring.

However, GPS is not the answer to every survey problem. The following difficulties may arise:
(1) A good electronic view of the sky is required so that the satellites may be ‘seen’ and ‘tracked’. There
should not be obstructions that block the ‘line of sight’ from the receiver to the satellite. This is usually not
a problem for the land surveyor but may become one for the engineering surveyor as a construction rises
from the ground. Satellite surveying cannot take place indoors, nor can it take place underground.
(2) The equipment concerned is expensive. A pair of GPS receivers costs about the same as three or four
total stations, though this will vary from manufacturer to manufacturer. Like total stations, however, prices
are falling while capabilities are increasing.
(3) Because satellites orbit the whole Earth, the coordinate systems that describe the positions of
satellites are global rather than local. Thus, if coordinates are required in a local datum or on a projection,
then the relationship between the local projection and datum, and the coordinate system of the satellite,
must also be known.
(4) The value of height determined by satellite is not that which the engineering surveyor would
immediately recognize. Since the coordinate system of GPS is Earth mass centred, then any height of a

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point on the Earth’s surface will be relative to some arbitrarily defined datum, such as the surface of an
ellipsoid. If height above the geoid (or mean sea level) is required, then the separation between the geoid
and the chosen ellipsoid will also be required. Some GPS receivers may have a geoid model in their
software to solve this problem; however, the model may be coarse.

GPS SEGMENTS
The GPS system can be broadly divided into three segments: the space segment, the control segment
and the user segment.

The space segment

This is composed of satellites weighing about 400 kg and powered by means of two solar panels with
three back-up, nickel-cadmium batteries

The operational phase consists of 28 satellites, at the present time, with three spares. They are in near-
circular orbits, at a height of 20 200 km above the Earth, with an orbit time of 12 hours (11h 58 min).

The six equally spaced orbital planes are inclined at 55° to the equator, resulting in five hours above the
horizon. The system therefore guarantees that at least four satellites will always be in view.
Each satellite has a fundamental frequency of 10.23 MHz and transmits two L-band radio signals.

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Modulated onto these signals are a Coarse Acquisition (C/A) code, now referred to as the Standard S-
code, and a Precise P-code. The L1 frequency has both the P- and S-codes, whereas the L2 has only
the P-code. The codes are pseudo-random binary sequences transmitted at frequencies of 1.023 MHz
(S-code) and 10.23 MHz (P-code).
The P-code provides what is termed the precise positioning service (PPS) and the S-code the standard
positioning service (SPS).
The SPS will provide absolute point position to an accuracy of 100–300 m; the PPS to an accuracy of 5–
10 m.
The codes are, in effect, time marks linked to ultra-accurate clocks (oscillators) on board the satellites.

The control segment

This has the task of supervising the satellite timing system, the orbits and the mechanical condition of the
individual satellites. Neither the timing system nor the orbits are sufficiently stable to be left unchecked for
any great period of time.
The satellites are currently tracked by five monitor stations, situated in Kwajalein, Hawaii, Ascension and
Diego Garcia, with the master control in Colorado Springs.
As the basic principle of position fixing using GPS is that of a resection, using distances to three known
points (satellites), the position of the satellites (in a known coordinate system) is critical.
The position of the satellite is obtained from data broadcast by the satellite and called the ‘broadcast
ephemeris’. The positional data from all the tracking stations are sent to the master control for processing.
These data, combined with the satellite’s positions on previous orbits, make it possible to predict the
satellite’s position for several hours ahead. This information is uploaded to the satellite, for subsequent
transmission to the user, every eight hours. Orbital positioning is currently accurate to about 10 m, but
would degrade if not continuously updated. The master control is also connected to the time standard of
the US Naval Observatory in Washington, DC. In this way, satellite time can be synchronized and data
relating it to Universal Time transmitted. Other data regularly updated are the parameters defining the
ionosphere, to facilitate the computation of refraction corrections to the distances measured.
The user segment
This consists essentially of, a portable receiver/processor with power supply and an omnidirectional
antenna

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The processor is basically a microcomputer containing all the software for processing the field data.

SATELLITE ORBITS
The German astronomer Johannes Kepler (1571–1630) established three laws defining the movement of
planets around the sun, which also apply to the movement of satellites around the Earth.
(1) Satellites move around the Earth in elliptical orbits, with the centre of mass of the Earth situated at one
of the focal points G. The other focus G_ is unused. The implications of this law are that a satellite will at
times be closer to or further away from the Earth’s surface depending upon which part of its orbit it is in.
GPS satellite orbits are nearly circular and so have very small eccentricity.

(2) The radius vector from the Earth’s centre to the satellite sweeps out equal areas at equal time
intervals (Figure 9.6). Therefore a satellite’s speed is not a constant. The speed will be a minimum when
the satellite is at apogee; at its furthest from the centre of the Earth and a maximum when it is at perigee,
the point of closest approach.

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(3) The square of the orbital period is proportional to the cube of the semi-major axis a, i.e.
. The value of the constant was later shown by Newton to be ⁄ where μ is the Earth’s

gravitational constant and is equal to 398 601 km3 s−2. Therefore ⁄ . So, whatever the

satellite’s orbital eccentricity, providing the semi-major axis is the same, then so will be the period.
Therefore these laws define the geometry of the orbit, the velocity variation of the satellite along its orbital
path, and the time taken to complete an orbit.

BASIC PRINCIPLE OF POSITION FIXING

The principle involves the measurement of distance (or range) to at least three satellites whose X, Y and
Z position is known, in order to define the user’s Xp, Yp and Zp position.
In its simplest form, the satellite transmits a signal on which the time of its departure (tD) from the satellite
is modulated. The receiver in turn notes the time of arrival (tA) of this time mark. Then the time which it
took the signal to go from satellite to receiver is (called the delay time). The measured range
R is obtained from ( )
Where c = the velocity of light.
Whilst the above describes the basic principle of range measurement, to achieve it one would require the
receiver to have a clock as accurate as the satellite’s and perfectly synchronized with it.
As this would render the receiver impossibly expensive, a correlation procedure, using the pseudo-
random binary codes (P or C/A), usually ‘C/A’, is adopted.

APPLICATIONS
The basic application of GPS in engineering surveying is to;
Establish control surveys, topographic surveys and setting-out on site.
Actually the three-dimensional spatial data normally captured using conventional surveying techniques
with a total station can be done by GPS, even during the night, provided sufficient satellites are visible.
On a national scale, horizontal and, to a certain extent, vertical control, used for mapping purposes and
established by classical triangulation with all its built-in scale error, are being replaced by three-
dimensional GPS networks. In Ordnance Survey, this has been dealt with in previous pages. The great
advantage of this to the engineering surveyor is that, when using GPS on a local level, there is no
requirement for coordinate transformation and the resultant plans are more consistent. Also, in mapping
at a local level there is, in effect, no need to establish a control network as it already exists in the form of
the orbiting satellite. Thus, time and money are saved.
Whilst the above constitutes the main area of interest for the engineering surveyor, other applications
will be briefly mentioned to illustrate the power and versatility of GPS.
Machine guidance (Figures 7.25 and 7.26)

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Earthmoving and grading plant are now being controlled in three dimensions using GPS in the realtime
kinematic (RTK) mode. Tests carried out at the IESSG (Nottingham University) using one antenna on the
cab and one at each end of the bulldozer blade gave accuracies of a few millimetres.
Combined with in-vehicle digital ground and design models, the machine operator can complete the
design without reliance on extraneous equipment such as sight rails, batter boards or lasers.
The advantage of GPS over existing systems that use total stations or lasers is that;
o Full three - dimensional information; is supplied to the operator permitting alignment positioning
as well as depth excavation or grading.
o Also, the system permits several items of plant to work simultaneously and over distances of 10
km from the base station. The plant is no longer dependent on the use of stakes, profile boards,
strings, etc., and so does not suffer from downtime waiting when ‘wood’ has been disturbed and
needs replacing. In this way the dozers, etc., can be kept running continuously with resultant
productivity gains.
Global mean sea level
The determination of MSL on a global scale requires a network of tide gauges throughout the world
connected to a single global reference frame. GPS is being used for this purpose within an international
programme called GLOSS (global level of the sea surface) established by the Intergovernmental
Oceanographic Commission. GPS is also involved in a similar exercise on a European basis.
Plate tectonics
Plate tectonics is centred, around the theories of continental drift and is the most widely accepted model
describing crustal movement. GPS is being used on a local and regional basis to measure three-
dimensional movement. On the local basis, inter-station vectors across faults are being continually
monitored to millimetre accuracy, whilst, on a regional basis, GPS networks have been established

‘Site vision GPS’ machine guidance by Trimble, showing 2 GPS antennae and the in-cab control system
on all continental plate boundaries. The information obtained adds greatly to the study of earthquake
prediction, volcanoes and plate motion.

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In a secondary way, it is also linked to global reference systems defined by coordinate points on the
Earth’s surface. As the plates continue to move, coordinate points will alter position and global reference
systems will need to be redefined.
Geographical information systems (GIS)
GIS has already been defined in the earlier chapters. Such is the growth in this area of spatial information
management that GPS systems have been specifically designed with GIS in mind.
The palm-size computer contains GIS Data PRO Office Software that converts GPS position information
to vector
GIS format, and a coding facility that permits smooth transfer of data to plan or computer database.
Navigation
GPS is now used in all aspects of navigation.
GPS voice navigation systems are now built in to several models of car; simply typing the required
destination into the on-board computer results in a graphical display of the route, along with voiced
directions. Similar systems are used by private boats and aeroplanes, whilst hand-held receivers are now
standard equipment for walkers and cyclists. A wristwatch produced by Casio contains a GPS system
giving position and route information to the wearer.
GPS can be used for fleet management when the position and status of vehicles can be transmitted to a
central control, thereby permitting better management of the vehicles, whilst the driver can use it as an
aid to route location.
It is used by surveying ships for major offshore hydrographic surveys. Ocean-going liners use it for
navigation purposes, whilst most harbours have a DGPS system to enable precise docking.
At the present time, aircraft landing and navigation are controlled by a variety of disparate systems. GPS
is gradually being introduced and will eventually provide a single system for all aircraft operations.
The uses to which GPS can be put are limited only by the imagination of the user. They can range from
the complexities of measuring gravity waves to the simplicity of spreading fertilizer in precision farming,
and include such areas of study as meteorology oceanography, geophysics and in-depth analysis on a
local and global basis. As GPS equipment and procedures improve, its applications will continue to grow.

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