Area and Volume Survey Engineering (RZ)

ENGINEERING SURVEY C 2005 / 2 /
AREA AND VOLUME
OBJECTIVES
General Objective : To know and understand the basic concepts of Area and Volume
Calculation
Specific Objectives : At the end of the unit you should be able to :-
 Explain the basic concept of Area and Volume Method.
 Define the usage of Area And Volume Calculation.
 Describe the methods that have been used in Area and
Volume Calculation .
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INPUTINPUT
2.1 INTRODUCTION
Estimation of area and volume is basic to most engineering schemes such as route
alignment, reservoirs, construction of tunnels, 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. Area may be required in connection with the purchase or sale of land,
with the division 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 practical.
• to enable contract estimates of time and cost to be made for proposed work.
• to form the basis of payment for work carried out.
It is frequently necessary as part of engineering surveying projects to determine the area
enclosed by the boundaries of a site or the volume of earthwork required to be moved.
Many of the figures involve accepted mensuration formulae (see 1.6 ) but it is more
common to meet irregular shapes and these require special attention.
2.2 PLAN AREAS
The basic unit of area in SI units is the square metre (m²) but for large areas the hectare is
a derived unit.
1 hectare (ha) = 10 000 m² = 2.471 05 acres
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2.2.1 Conversion Of Plannimetric Area Into Actual Area
Let the scale of the plan be 1 in H (or as representative fraction 1/H). Then
1 mm is equivalent to H mm and 1 mm² is equivalent to H² mm² is
equivalent to H mm², i.e.
H² x 10-6
m²
2.3 AREA CALCULATION
Areas of ground may be obtained from the plotted plan but results are only as accurate as
it is possible to scale off the drawings. Accuracy is greatly increased by using the
measurements taken in the field. In most surveys the area is divisible into two parts :
a) The rectilinear areas enclosed by the survey lines
b) The irregular areas of the strips between these lines and the boundary
In order to calculate the area of the whole, each of these areas must be evaluated
separately because each is defined by a different form of geometrical figure.
2.3.1 Rectilinear Areas
The method of evaluating the rectilinear area enclosed by survey lines
depends on the method of survey.
a) If chain surveying is used, the areas of the triangles forming the survey
network are calculated from the field dimensions from the formula :
Area = √ s(s – a) (s – b) (s – c)
Where a, b and c = the lengths of the triangles sides and
s = (a + b + c) / 2
b) If traversing is used and the survey stations are coordinated, the computed
coordinated are used in the area calculation.
Whichever calculation method is used, checks must be applied to prove
the area calculations. In a chain survey network the work must be arranged so that
two different sets of the triangles forming the rectilinear figure are used in
evaluating the total area, which is thus twice calculated. These two results will not
normally agree precisely because the network will not be geometrically perfect.
Owing to observational errors, the two results are meaned to produce the final
rectilinear area. When areas are calculated from coordinates, the calculation must
be repeated another way to prove the result.
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2.3.2 Irregular Areas
Unless boundaries are straight and the corner points coordinated there are
usually irregular strips of ground between the survey lines and the property
boundaries. The area of the irregular strips are either positive or negative to the
rectilinear area and since they are divided up by offsets between which the
boundary is supposed to run straight, they are computed as a series of trapezoids.
The mean of each pair of offsets is taken and multiplied by the chainage between
them. Where the offsets are taken at regular intervals, the trapezoidal rule or
Simpson’s rule for areas is used, (see section 2.6).
NOTE
a. The field work should be arranged to overcome difficulties with corners.
This is usually achieved by extending the survey line to the boundary,
allowing for the triangular shape which may occur.
b. In order to check the irregular area the calculations should be repeated
by another person, or a check against gross error may be made taking out
a planimeter area of the plot.
2.4 CALCULATING AREA FROM A CHAIN SURVEY
The figure shows the rectilinear area ABCD, which is calculated first. Their
regular strips between the chain lines and the boundary must be separately evaluated and
either added or subtracted as necessary from the main rectilinear area calculation result.
The following data were obtained from the chain survey of the site :
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AB - 63.0 m
BC - 45.0 m
CD - 60.0 m
DA - 78.0 m
BD - 93.3 m
AC - 76.0 m
SOLUTION
The rectilinear area from A = √ ((s – a) (s – b) (s – c))
The area of triangle ACD = √(107(31) (47) (29))
= 2126.3 m2
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Chainage AD Offset
A 0.0 0.0
16.0 6.0
33.0 7.0
40.0 0.0
49.0 7.0
61.0 7.0
68.0 0.0
B 78.0 11.0
89.0 5.0
93.0 9.0
Chainage CD Offset
C 0.0 0.0
10.0 4.2
20.0 6.4
30.0 8.1
40.0 10.3
50.0 11.3
D 60.0 13.2
AB and BC are straight boundaries. Offsets to the
irregular boundaries are as follows :

The area of triangle ABC = √(92(29) (47) (16))
= 1416.4 m2
Area of ABCD = 2126.3 + 1416.4
= 3542.7 m2
Check :
The area of triangle ABD = √((117.15 (54.15) (39.15) (23.85))
= 2433.8 m2
The area of triangle ABD = √(( 99.15 (39.15) (54.15) (5.85)))
= 1108.9 m2
Area of ABCD = 2433.8 + 1108.9
= 3542.7 m2
Area of triangle ABD: Plus Minus
(0+6) x 2 x 16 = 48.0
(6+7) x 2 x 17 = 110.5
(7+0) x 2 x 7 = 24.5
(0+7) x 2 x 9 = 31.5
(7+7) x 2 x 12 = 84.0
(7+0) x 2 x 7 = 24.5
(0+11) x 2 x 10 = 55.0
(11+9) x 2 x 15 = 150.0
388.5
140.0
- 140.0
248.5 m2
(total plus area on AD)
2.5 CALCULATING AREAS
FROM COORDINATES
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A = Area
SPECIMEN QUESTION
Calculate the area of the figure ABCDEF of which the coordinates are listed
below.
SOLUTION
The calculation is tabulated as shown :
Product
Station Easting Northing
E + E
Double
Longitude
ΔN
A 150 100
B 95.2 164.3 245.2 64.3 15 766.36
C 127.9 210.7 223.1 46.4 10 351.84
D 176.3 239.8 304.2 29.1 8 852.22
E 219.4 222.4 395.7 -17.4 6 885.18
F 237.5 163.8 456.9 -58.6 26 774.34
A 150 100 387.5 -63.8 24 722.50
34 970.42 58 382.02
34 970.42
2A = 23 411.60
Area = 11 705.8 m2
= 1.1706 ha
2.6 AREAS OF IRREGULAR FIGURES
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There are several practical situations where it is necessary to estimate the area of
irregular figures. Examples include estimation of areas of plots of land by surveyors,
areas of indicator diagrams of steam engines by engineers and areas of water planes and
transverse sections of a ship by naval architects. There are many methods whereby the
area of an irregular plane surface may be found and these include:
(a) Use of a planimeter,
(b) Trapezoidal rule,
(c) Mid-ordinate rule and
(d) Simpson’s rule.
2.6.1 The planimeter
A planimeter is an instrument for directly measuring areas bounded by an
irregular curve. There are many different types of the instrument but all consist
basically of two rods AB and BC, hinged at B (see Fig. 2.1). The end labelled A is
fixed, preferably outside of the irregular area being measured. Rod BC carries at
B a wheel whose plane is at right angles to the plane formed by ABC. Point C,
called the tracer, is guided round the boundary of the figure to be measured. The
wheel is geared to a dial which records the area directly. If the length BC is
adjustable, the scale can be altered and readings obtained in mm2
, cm2
, m2
and so
on.
FIGURES 2.1 : Planimeter
(Source : Mathematics for
Technicians, S. Adam)
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2.6.2 Trapezoidal rule
To find the area ABCD in Fig. 2.2, the base AD is divided into a number
of equal intervals of width d. This can be any number; the greater the number the
more accurate the result. The ordinates y1, y2, y3, etc. are accurately measured.
The approximation used in this rule is to assume that each strip is equal to the area
of a trapezium.
FIGURE
2.2 : Trapezoidal rule (Source : Mathematics for Technicians, S. Adam)
The area of a trapezium = ½ (sum of the parallel sides) (perpendicular distance
between the parallel sides).
Hence for the first strip, shown in Fig. 2.2, the approximate area is ½ (y1 + y2)d.
For the second strip area is ½ (y1 + y2)d and so on. Hence the approximate area of
ABCD = ½ (y1 + y2)d + ½ (y3 + y4)d + ½ (y3 + y4)d + ½ (y4 + y5)d
+ ½ (y5 + y6)d + ½ (y6 + y7)d
= ½ y1 d + ½ y2 d + ½ y2 d + ½ y3 d + ½ y3 d + ½ y4 d + ½ y4 d
+ ½ y5 d + ½ y5 d + ½ y6 d + ½ y6 d + ½ y7 d
= ½ y1 d + ½ y2 d + ½ y3 d + ½ y4 d + ½ y5 d + ½ y6 d + ½ y7 d
= d [ ( y1 + + y7 ) / 2 + y2 + y3 + y4 + y5 + y6 ]
Generally, the trapezoidal rule states that the area of an irregular figure is given
by:
Area = (width of internal) [½ (first + last ordinate) + sum of remaining ordinates]
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2.6.3 Mid-ordinate rule
FIGURE 2.3 : Mid-ordinate rule method
(Source : Mathematics for Technicians, S. Adam)
To find the area of ABCD in Figure 2.3 the base AD is divided into any
number of equal strips of width d. (As with the trapezoidal rule, the greater the
number of intervals used the more accurate the result.) If each strip is assumed to
be a trapezium, then the average length of the two parallel sides will be given by
the length of a mid-ordinate, i.e. an ordinate erected in the middle of each
trapezium. This is the approximation used in the mid-ordinate rule.
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The mid-ordinates are labelled y1, y2, y3, etc. as in Fig. 18.3 and each is then
accurately measured. Hence the approximate area of ABCD
= y1 d + y2 d + y3 d + y4 d + y5 d + y6 d
= d (y1 + y2 + y3 + y4 + y5 + y6 )
where d = ( length of AD / number of mid-ordinates )
Generally, the mid-ordinate rule states that the area of an irregular figure is given
by:
2.6.4 Simpson’s rule
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Area = (width of interval) (sum of mid-
ordinates)

FIGURE 2.4 : Simpson’s rule (Source : Mathematics for Technicians, S. Adam)
To find the circa A BCD in Figure 2.4 the base AD must be divided into
an even number of strips of equal width d. Thus producing an odd number of
ordinates. The length of each ordinate, y1, y2, y3, etc., is accurately measured.
Simpson's rule states that (the area of the irregular area ABCD is given by;
Area of ABCD = d / 3 [(y1 + y7 ) + 4(y2 + y4 + y6) + 2(y3 + y5)]
More generally, the calculation of the area of:
Area = 1/3 (width of interval) [(first and last ordinates) + 4( sum of even ordinates)
+ 2 (sum of remaining odd ordinates)]
When estimating areas of irregular figures, Simpson's rule is generally regarded as
the most accurate of the approximate methods available.
Activity 2a
2.1 The values of the y ordinates of a curve and their distance x from the origin are
given in the table below. Plot the graph and find the area under the curve by :
x 0 1 2 3 4 5 6
y 2 5 8 11 14 17 20
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a) The trapezoidal rule
b) The mid-ordinate rule
c) Simpson’s rule
2.2 Sketch a semicircle of radius 10cm. Erect ordinates at intervals of 2 cm and
determine the lengths of the ordinates and mid-ordinates. Determine the area of the
semicircle using the three approximate methods. Calculate the true area of the
semicircle.
Feedback 2a
2.1)
FIGURE 2.5 : Graph of y against x
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Try your
best to
answer this
question.

a) Trapezoidal rule
Using 7 ordinates with interval width of 1 the area under the curve is:
Area = 1 [ ½ (2 + 20) + 5 + 8 + 11 + 14 + 17 ]
= [ 11 + 5 + 8 + 11 + 14 + 17 ]
= 66 square units
b) Mid-ordinate rule
Using 6 intervals of width 1 the mid-ordinates of the 6 strips are measured. The area
under the curve is:
Area = 1 (3.5 + 6.5 + 9.5 + 12.5 + 15.5 + 18.5)
= 66 square unit
c) Simpson’s rule
Using 7 ordinates, given an even number of strips, i.e. 6, each of width 1, thus the area
under the curve is:
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Area = 1 / 3 [ (2 + 20) + 4(5 + 11 + 17) + 2 (8 + 14) ]
= 1 / 3 [ 22 + 4(33) + 2(22)]
= 1 / 3 [ 22 + 132 + 44 ]
= 198 / 3
= 66 square units
The area under the curve is a trapezium and may be calculated using the formula
½(a+b)h, where a and b are the lengths of the parallel sides and h the perpendicular distance
between the parallel sides.
Hence area = ½(2 + 20)(6) = 66 square units. This problem demonstrates the
methods for finding areas under curves. Obviously the three 'approximate' methods would
not normally be used for an area such as in this problem since it is not 'irregular'.
2.2). The semicircle is shown in Fig. 2.6 with the lengths of the ordinates and mid-
ordinates marked, the dimensions being in centimetres.
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FIGURE 2.6 : Sketch a semicircle
a) Trapezoidal rule
Area = 2 [ ½ (0 + 0) + 6.0 + 8.0 + 9.15 + 9.80 + 10.0 + 9.80 + 9.15 + 8.0 + 6.0 ]
= 2 (75.90)
= 151.8 square units
b) Mid-ordinate rule
Area = 2 [ 4.3 + 7.1 + 8.65 + 9.55 + 9.95 + 9.95 + 9.95 + 8.65 + 7.1 + 4.3 ]
= 2 (79.10)
c) Simpson’s rule
Area = 2/3 [ (0 + 0) + 4(6.0 + 9.15 + 10.00 + 9.15 + 6.00) + 2(8.0 + 9.8 + 8.0)]
= 2/3 [0 + 4(40.3) + 2(35.6)]
= 2/3 (161.2 + 71.2)
= 2/3 (232.4)
The true area is given by π r² / 2, i.e π (10)² / 2 = 157.1 square units
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Well done!
Keep it up!.

INPUTINPUT
2.7 VOLUME CALCULATION
In construction works, the excavation, loading, hauling and dumping of earth
frequently forms a substantial part of the project. Payment must be made for the labour
and plan needed for earthworks and this is based on the quantity or volume handled.
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These volumes must be calculated and depending on the shape of the site, this may be
done in three ways :
i) by cross-sections, generally used for long, narrow works such as roads,
railways, pipelines, etc.
ii) by contours, generally used for larger areas such as reservoirs, landscapes,
redevelopment sites, etc.
iii) by spot height, generally used for small areas such as underground tanks,
basements, building sites, etc.
2.8 CROSS SECTION VOLUME CALCULATION
Cross-sections are established at some convenient intervals along a centre line of
the works. Volumes are calculated by relating the cross-sectional areas to the distances
between them. In order to compute the volume it is first necessary to evaluate the cross-
sectional areas, which may be obtained by the following methods:
i) by calculating from the formula or from first principles the standard cross-
sections of constant formation widths and side slopes.
ii) by measuring graphically from plotted cross-sections drawn to scale, areas
being obtained by plannimeter or division into triangles or square.
NOTE :
The graphic measure of the cross-sectional area is most often used and provides a
sufficiently accurate estimate of volume, but for railways, long embankments,
breakwaters, etc., with fairly regular dimensions, the use of formulae may be easier and
perhaps more accurate.
2.8.1 Prismoidal Method
In order to calculate the volume of a substance, its geometrical shape and
size must be known. A mass of earth has no regular geometrical figure in most
approaches. The prismoid is a solid, consisting of two ends which form plane,
parallel figures, not necessarily of the same number of sides and which can be
measured as cross-sections. The faces between the parallel ends are plane surfaces
between straight lines which join all the corners of the two end faces. A prismoid
can be considered to be made up of a series of prisms, wedges and pyramids, all
having a length equal to the perpendicular distance between the parallel ends. The
geometrical solids forming the prismoid are described as follows :
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i) Prism, in which the end polygons are equal and the side faces are
parallelograms.
ii) Wedge, in which one end is a line, the other end a parallelogram, and the
sides are triangles and parallelograms.
iii) Pyramids, in which one end is a point, the other end a polygon and the
side faces are triangles.
The Prismoidal Formula
Let D = the perpendicular distance between the parallel end planes,
A1 and A2 = the areas of these end planes,
M = the mid-area, the area of the plane parallel to the end planes
and midway between them,
V = the volume of the prismoid and
a1, a2, m, v = the equivalent for any prism, wedge or pyramid forming the
prismoid
then in a prism a1, = a2, = m
and in a wedge a2 = 0 and m = 1/2 a1
and in a pyramid a2 = 0 and m = 1/4 a1
Prism volume v = D . a1 = D/6 (6 . a1 ) = D/6 (a1, + 4m + a2)
Wedge volume v = ½ D . a1 = D/6 (3 . a1 ) = D/6 (a1, + 4m + a2)
Pyramid volume v = 1/3 D . a1 = D/6 (2 . a1 ) = D/6 (a1, + 4m + a2)
As the volume of each part can be expressed in the same terms, the volume of the
whole can take the same form. Thus the prismoidal formula is expressed in the
following way :
V = D/6 (A1, + 4M + A2)
Note :
A. M does not represent the mean of the end areas A1 and A2 except where
the prismoid is composed of prisms and wedges only.
B. The formula gives the volume of one prismoid of which the end and mid-
sectional areas are known.
The prismoidal formula may be used to calculate volume if a series of cross-
sectional areas, A1, A2, A3,…. An, have been established a distance d apart. Each
alternate cross-section may be considered to be the mid-area M of a prismoid of
length 2d.
Then the volume of the first prismoid of length 2d :
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= 2d / 6 (A1, + 4A2 + A3)
and of the second = 2d / 6 (A3, + 4A4 + A5)
and of the nth = 2d / 6 (An-2, + 4An-1 + An)
summing up the volumes of each prismoidal :
V = d / 3 (A1, + 4A2 + 2A3 + 4A4…… + 2An-2 + 4An-1 + An)
Which is Simpson’s rule for volumes.
Specimen Question
Calculate, using the prismoidal formula, the cubic contents of an embankment of
which the cross-sectional areas at 15m intervals are as follows :
Distance (m) 0 15 30 45 60 75 90
Area (m2
) 11 42 64 72 160 180 220
Solution,
V = 15 / 3 (11 + 220 + 4 ( 42 + 72 + 180 ) + ( 64 + 160))
V = 5 ( 231 + 1176 + 448 )
V = 9275 m3
Note :
A. The 15m interval is divided by 3, as the length of the individual prismoids
used is 30m, which in the prismoidal formula is divided by 6.
B. A mass of earth, length double the usual cross-sectional interval of 15m,
20m or 25m, is considerably different from a true prismoid, so this method
is not as accurate as it would be if the true mid-sectional area had been
measured. This results in the use of prismoids of length equal to, instead
of double, the interval between cross-sections.
2.8.2 End Areas Method
It is no more accurate to use the prismoidal formula where the mid-
sectional areas have not been directly measured than it is to use the end areas
formula, particularly as the earth solid is not exactly represented by a prismoid.
Using the same symbols the volume may be expressed as :
v = d [ ( A1 + A2) / 2 ]
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although this is only correct where the mid-area is the mean of the end areas :
M = ( A1 + A2) / 2
However, in view of the inaccuracies that arise in assuming any geometric
shape between cross-sections and because of bulking and settlement and the fact
that the end areas calculation is simple to use, it is generally used for most
estimating purposes.
Note :
A. The summation of a series of cross-sectional areas by this method
provides a total volume :
V = d{[( A1 + A2) / 2 ] + A2 + A3 + …… An-1}
Specimen Question
Calculate, using the end areas method, the cubic contents of the embankment of
which the cross-sectional areas at 15m intervals are as follows :
Distance (m) 0 15 30 45 60 75 90
Area (m2
) 11 42 64 72 160 180 220
Solution
V = 15{[ (11 + 220) / 2 ] + 42 + 64 + 72 + 160 + 180 }
V = 9502.5 m3
2.9 VOLUME CALCULATION FOR CONTOUR LINES
Contour lines may be used for volume calculations and theoretically this is the
most accurate method. However, as the small contour interval necessary for accurate
work is seldom provided due to cost, high accuracy is not often obtained. Unless the
contour interval is less than 1m or 2m at the most, the assumption that there is an even
slope between the contour is incorrect and volume calculation from contours become
unreliable.
The formula used for volume calculation is the end areas formula of Simpson’s
rule for volumes, the distance d in the formula being contour interval. The area enclosed
by each contour line is measured, usually by plannimeter, and these areas A1, A2, etc.,
are used in the formula as before ( see the end areas method). If the prismoidal method is
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used, each alternate contour line is assumed to enclose a mid-area or the outline of the
mid-area can be interpolated between the existing contour intervals.
This illustration shows an area
contoured at 5m intervals and how the
contours of proposed works, in this
case a dam wall with an access road
through a cutting, shown as packed
lines, define the plan outline of the
works. This also allows the volume of
the earthworks to be calculated using
the positions of the contour lines.
Picture 4.1 : Intersection Of
Contoured Surfaces (Source : Land
Survey, Ramsay)
The volume of the dam wall and the amount of cut may be obtained from the
contour lines by calculating the volume of ground within the working area down to a
common level surface and then calculating the new volume from the formation contour
lines, the difference being the change in volume due to the works. This volume
calculation is more usually carried out by using the cross-sectional method. The use of
contours is a practical method of calculating volumes in several cases, one of which
being the calculation of water at various levels in a reservoir. For example, in picture 4.1
the volume of water which could be contained up to the level of the 60m, contours could
be calculated as follows from these data :
Contour above datum (m) 50 52.5 55 57.5 60
Area (m2
) 12 135 660 1500 1950
Using the end areas method :
V = 2.5 { [ (12 + 1950)/2] + 135 + 660 + 1500 }
= 8190 m3
Using Simpson’s rule from the prismoidal formula :
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V = ( 2.5 / 3 ) [ 12 + 1950 + 4(135 + 1500) + 2 (660) ]
= (2.5 / 3 ) (9822)
= 8185 m3
Note :
The small volume of water below 50m (not included in the above calculation) would be
estimated from the interpolated depth of 2m at the deepest point, using the end areas
formula, the lowest end area being 0, thus :
V = [ ( 12 + 0 ) / 2 ] x 2
V = 12m3
This would then be added to either of the results above.
2.10 VOLUME CALCULATION FROM SPOT HEIGHT
This is a method of volume calculation frequently used on excavations where
there are vertical sides covering a fairly large area, although it can be used for excavation
with sloping sides. The site is divided into squares or rectangles, and if they are of equal
size the calculations are simplified. The volumes are calculated from the product of the
mean length of the sides of each vertical truncated prism ( a prism in which the base
planes are not parallel ) and the cross-sectional area. The sizes of the rectangles is
dependent on the degree of accuracy required. The aim is to produce areas such that the
ground surface within each can be assumed to be plane.
Specimen Question
Picture 4.2 shows the reduced levels of a rectangular plot which is to be excavated to a
uniform depth of 8m above datum. Calculate the mean level of the ground and the
volume of earth to be excavated.
Note :
A. The mean or average level of the
ground is that level of ground which
would be achieved by smoothing the
ground off level, assuming that no
bulking would take place.
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B. The mean level of the ground is the mean of the mean height of each
prism. It is not the mean of all the spot heights.
Picture 4.2 : Calculating volume from spot height
on a levelling grid. (Source : Land Survey, Ramsay)
Solution
(a) Calculation from rectangles :
Station R.L.
Number of times
the Product
R.L. is used = n (R.L.) x n
A 12.16 1 12.16
B 12.48 2 24.96
C 13.01 1 13.01
D 12.56 2 25.12
E 12.87 4 51.48
F 13.53 2 27.06
G 12.94 1 12.94
H 13.27 2 26.54
J 13.84 1 13.84
∑n = 16 207.11
Mean level = 207.11 / 16
= 12.944 m
Depth of excavation = 12.944 – 8.00
= 4.944
Volume = Total area x Depth
= 30 x 20 x 4.944
= 2966.4 m3
(b) Calculation from triangles
It is usually more accurate to calculate from triangles as the upper base of the triangular
prism is more likely to correspond with the ground plane than the larger rectangle. The
mean level of each prism is then the mean of the three height enclosing the triangle
instead of four as before.
Station R.L.
Number of times the Product
R.L. is used = n (R.L.) x n
A 12.16 1 12.16
B 12.48 3 37.44
C 13.01 2 26.02
D 12.56 3 37.44
E 12.87 7 90.09
F 13.53 2 27.06
G 12.94 2 25.88
H 13.27 2 26.54
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J 13.84 2 27.68
∑n = 24 310.55
Mean level = 310.55 / 24
= 12.940 m
Depth of excavation = 4.960
Volume = 30 x 20 x 4.944
= 2966.4 m3
Note :
The diagonal forming the triangles would be noted in the field book on the grid layout to
conform most suitably with the ground planes.
Activity 2b
2.3 An embankment is to be formed with its centre line on the surface (in the form of a
plane) on full dip of 1 in 20. If the formation width is 12.00m and the formation
heights are 3.00m, 4.50m and 6.00m at intervals of 30.00m, with side slopes 1 in 2,
calculate the volume between
the end sections.
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Calculate
a). Volume by mean areas
b). Volume by end areas
c). Volume by prismoidal rule
2.4 Given the previous example but with the centre line turned through 90º, calculate
volume
a) By mean areas
b) By end areas
c) By prismoidal
Feedback 2b
2.3 Area (1) = h1 (w + mh1)
= 3.00 [ 12.00 + (2 x 3.00) ] = 54.00 m2
Area (2) = 4.50 [ 12.00 + (2 x 4.50) ] = 94.50 m2
Area (3) = 6.00 [ 12.00 + (2 x 6.00) ] = 144.00 m2
Volume :
26
26
Try your
best to
answer this
question.

a). By mean areas
V = W(A/n) = 60.00 ( 54.00 + 94.50 + 144.00 ) / 3
= 5850.0 m3
b). By end areas
V = w ( A1 + 2A2 + A3 ) / 2
= 30.00 (54.00 + 189.00 + 144.00) / 2
= 5805.0 m3
c). By Prismoidal Rule
V = w ( A1 + 4A2 + A3 ) / 3
= 30.00 (54.00 + 378.00 + 144.00)/3
= 5760.0 m3
2.4 A = m ( h²0 k² + w² / 4 + wh0 m) + wh0
( k² - m² )
Cross-sectional areas
A1 = 2 [ (3.00² x 20² ) + ( 0.25 x 12.00²) + ( 12.00 x 3.00 x 2 ) + (12 x 3.00)
( 20² - 2² )
= [ (3600.00 + 36.00 + 72.00) / 198 ] + 36.00 = 54.73 m²
A2 = [ ( 8100.00 + 36.00 + 108.00 ) / 198 ] + 54.00 = 95.64 m²
A3 = [ ( 14400.00 + 36.00 + 144.00 ) / 198 ] + 72.00 = 145.64 m²
Volume
a). By mean areas
V = 60.00 ( 54.00 + 95.64 + 145.64 ) / 3 = 5920.2 m³
b). By end areas
V = 30.00 ( 54.73 + 191.28 + 145.64 ) / 2 = 5874.8 m³
c). By prismoidal rule
V = 30.00 ( 54.73 + 382.56 + 145.64 ) / 3 = 5829.3 m³
27
27
Well done!
Keep it up!.

Self Assessment
Calculate the volumes in Figure 1 and Figure 2.
Figure 1
28
28

Feedback to Self Assessment
Figure 2
Figure 1
A1 = [ ( 0.75 + 4.75 ) / 2 ] x 7 = 19.25 ft²
A2 = [ ( 0.75 + 3.75 ) / 2 ] x 5 = 11.25 ft²
A3 = [ ( 0.75 + 2.75 ) / 2 ] x 3 = 5.25 ft²
Volume, V = L / 6 ( A1 + 4Am + A2 )
29
29
How ???

= 17 / 6 (19.25 + 4 x 11.25 + 5.25)
= 17 / 6 ( 19.25 + 45.00 + 5.25 )
= 17 / 6 ( 69.50 )
= 1181.50 / 6
= 196.92 ft³
= 7.29 yrd³
Figure 2
Volume, V = h / 3 ( area of base )
= 27.4 / 3 ( 13.5 x 13.5 )
= 1664.6 m³
30
30
Congratulations you can can
proceed to the next unit.
IN
P
U
T

Area and Volume Survey Engineering (RZ)

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