surveying

1. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 1
UNIT SIX AREA COMPUTATION
6.1 INTRODUCTION
One of the primary objects of cadastral surveys is to determine the area of land in plan. The land under consideration can
be the property of the person, or a site on which building is to be erected or to be used as a parking lot.
Its often necessary to compute the area of a tract of land which may be regular or irregular in shape. Land is frequently
bought and sold on the basis of cost per unit area.
The most common unit of area for lots is the m2 Large tracts such as huge tracts of forest, farm land etc are measured in
hectares (ha) or gashas’.
I hectar=10,000m2
1 gasha=40h
1 gasha=400,000m2
6.2 METHODS OF MEASURING AREA
There are different methods for measuring area. They are
1. Division of the tract or plot in to simple figures (i.e triangles, rectangles and trapeziums
2. Measuring offsets from a chain line which is also called a base line
3. Coordinate method
When the plan or map of an area is available, however irregular it may be,
planimeter can be run over the enclosing lines to compute the area of the plot
6.2.1 THE SIMPLE TRIANGLE
Where an area is triangular in shape or is made up of a series of triangles, the following formulae are used in
computing the area.
a) If the lengths of the three sides have been field measured or given , the following Heros formula can be used
to compute the area:
Area=√s(s-a)(s-b)(s-c)
Where a, b, c are sides of the triangle and S=1/2 (a+ b+ c)
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2. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 2
A
B
C
b
a
h
c
b) Two sides and the included angles are field measured or given : let us suppose that sides a, b and <c have
been measured.
Area= ½ base x height=1/2 bh but h=a sinc
Therefore , area=1/2 ba Sinc
C) Any side and three angles are field measured or given
Let us suppose that side a and the angles A,B,C have been measured.
Area of the Δ=1/2 ba sinC
Applying the law of sines to fig 1
b/ sinB=a/sinA then b=a sinB/sinA
by substituting the value of b, we have
Area of the Δ=1/2 (a sinB/ sinA) a sinC
=(a² sin B sinC) 2sinA
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3. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 3
Ex1 A chain survey provides the length of the sides of the closed polygon shown in fig 2.
Compute the total area of the plot of land.
B
C
D
E
A
25
18
25
A1
26
A2
32
A3
14
20
Soln
Let A1, A2, A3 denote the areas of the triangle as shown in fig 2, and S1, S2, S3 denote the half perimeters of the
triangles A1,A2 and A3 respectively.
Step1: computation of half perimeters
S1=1/2 (25+18+26)=34.5m
S2=1/2(26+25+32)=41.5m
S3=1/2 (32+20+14)=33m
Step2 : computation of areas
A=√s(s-a)(s-b)(s-c)
A1=√34.5(34.5-25)(34.5-18)(34.5-26)=214.40m2
in a similar manner , A2=317.54m2, A3=90.28m2
Total Area= A1+A2+A3
=622.22m2=0.06ha
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4. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
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6.2.2 BY TAKING OFFSETS FROM A BASE LINE
If the boundaries of a tract are irregular, its not possible to run the traverse along the boundaries.
The traverse is usually run at convenient distance from the actual boundaries. The offsets from the traverse to the
irregular boundary are taken at regular intervals or if necessary at irregular intervals. The area between the traverse line
and the irregular boundary is determined by
1) Average ordinate rule
2) Trapezoidal rule
3) Simpsons rule
1) Average ordinate rule
Figure 3 explains this method .
d d d d d d d
h8
h7
h6
h5
h4
h3
h2
h1
d
If h1,h2……….h8are the ordinates to the boundary from the baseline
Average ordinate = (h1+h2+……+h8)/8
And area=average ordinate x length
=L(h1+h2+……h8)/8
2. Trapezoidal rule
which is obtained by considering each part as a trapezium and then adding the part mass together. Area is
equal to product of the common interval d and sum of intermediate ordinates plus average of the first and last
ordinates. If the intervals are not equal the areas of the trapeziums have to be computed separately and added
together.
n-1
Area= d/2( he+ h’e +2 ∑ hi)
i=2
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5. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 5
Where d= distance between the offsets (m) , he and h’e= offsets at two ends (m), and
n-1
∑ hi= sum of the intermediate offsets (m)
i=2
3. Simpsons rule
In the rules stated above the irregular boundary consists of a number of straight lines. If the boundary is curved, it
can be approximated as a serious of straight lines. Alternatively, Simpsons rule is applied which assumes that the
short lengths of boundaries between the ordinates are parabolic arcs.
Since the trapezoidal rule presupposes very short intercepts along the base line. The area computed by this
method is less accurate than that computed by Simpsons rule for curved boundary.
In order to apply Simpsons rule for the computation of areas of tracts, two condition must be fulfilled. They are
i) There must be an even number of equal intercepts, say 8, along the base line i.e. the number of offsets
must be odd.
ii) The offsets must be at regular intervals
To get the area by simpson’s rule, add first and last ordinates to four times the even ordinates and two
time the odd ordinates and multiply the sum by one third the common interval.
Area=d/3(he+h’e+2∑hodd+4∑heven )
Where he, h’e=offset at the two ends
∑h odd=the sum of the odd offsets except the first and the last (the 3rd,5th,7th, etc.)
∑h even=the sum of the even offsets (the 2nd, 4th, 6th, etc.)
Ex2: the following offsets were taken from a chain line AB to a hedge
corner C D E F G H I J K L M
Offset h 1 h2 h3 h4 h5 h6 h7 h8 h9 h10 h11
Length, m 0 20 40 60 80 120 160 200 240 270 300
Calculate the area enclosed by the chain line, the hedge and the end offsets by
i) Simpsons rule, ii) trapezoidal rule
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6. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 6
C
D
E F G H I
J
K L
M
24m
h11
h10
h9
h8
h7
h6
h5
h4
h3
h2
h1
O 20 40 60 80 120 160 200 240 270 300
20m
16m
12m
8m
10m
14m
16m
20m
22m
26m
A
survey line
Soln
Let A be the required total area, A1 be the area of section C-G, A2 be the area of section G-K, and A3 be the area of
section K-M
1) by Simpsons rule
A1=20/3 (24+8)+2(16)+4(20+12)=1280m2
A2=40/3 (8+20)+2(14)+4(10+16)=2133.33m2
A3=30/3(20+26)+2(0)+2(0)+4(22)=1340m2
A=A1+A2+A3=4753.33m2
ii) by Trapezoid rule
A1=20/2(24+8)+2(20+16+12)=1280m2
A2=40/2(8+20)+2(10+14+16)=2160m2
A3=30/20(20+26)+2(22)=1350
A=A1+A2+A3=4790m2
6.2.3 Coordinate Method
In this method independent coordinates of the points are used in the computation of areas.
o B(XB,YB)
C(XC,YC)
D(XD,YD)
A(XA,YA)
Y
Total area of the traverse
A=1/2(Xc+XB)(Yc-Yb)+1/2 (Xd+Xc)(Yd-Yc)
-1/2 (XB+Xa)(Ya-Yb)-1/2(Xd+Xa)(Yd-Ya)
Or 2(A)=XaYb+YbYc+YcYd+XdYa-XbYa-XcYb-XdYc-XaYd
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7. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 7
=(XaYb-XbYA)+XbYc-XcYb)+(XcYd-XdYc)+(XdYa-XaYd)
The above relation can be expressed as follows for easy remembrance.
YA
YB
YC
YD
YA
XA
XB
XC
XD
XA
Two sums of the product should be taken.
i) Product of all adjacent terms taken down to the left, i.e.
XAYB,XBYC,XCYD,XDYA
ii) Product of all adjacent terms taken down to the right, i.e.
YAXB,YBXC, YCXD,YDXA
the traverse area is equal to half the absolute value of the difference between these two sums . in applying this procedure,
its to be observed that first coordinate listed must be repeated at the end of the list.
Ex3: using the following data, calculate the area of the closed traverse by the coordinate method.
Corner A B C D E F
X,m 300 800 1100 1300 900 300
Y,m 1000 1200 1000 600 200 500
Solution
Corner A B C D E F A
X,m 300 800 1100 1300 900 300 300
Y,m 1000 1200 1000 600 200 500 1000
i) Let S1 be the sum of the product of adjacent diagonal terms taken down to the right.
S1=300x1200+800x1000+1100x600+1300x200+900x500+300x1000=2,830,000
ii) Let S2 be the sum of the product of adjacent diagonal terms taken down to the left
S2=800x1000+1100x1200+1300x1000+900x600+300x200+300x500=4,170,000
The algebraic summation S of these two terms is
S=2,830,000-4,170,000
=-1,340,000
There for Area=1/2(1,340,000=670,000m²=67ha
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8. Addis Ababa University Ethiopian Institute of Architecture, Building construction and City Development
AREA COMPUTATION
Compiled by Ebisa Tesfaye Page 8
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