COMPUTATION OF AREA
INTRODUCTION:
The term ‘area’ in
surveying refers to the area of a tract of land projected upon the horizontal
plane, and not to the actual area of the land surface.
The following are
some unit in which area in expressed – Square Meters (Sqm.), Hectares (1
hectare = 10,000 Sqm.), Square Feet, Acres (1 acre = 43.560 Sq. ft.)
METHODS OF
COMPUTATION OF AREA:
Generally, area can
be calculated from the from the following method –
§
COMPUTATION AREA FROM
FIELD NOTES:
Step – 01
Ø During the
cross-stuff survey, the area of the field can be directly calculated from field
notes.
Ø During the survey wok
the whole area is divided into some geometrical figures, such as squares,
rectangles, triangles, and trapeziums, and then the area is calculating by
using the following formulas –
a) Area of Squares –
Area = a2
Where ‘a’ is the Sides of the Square
a) Area of Triangle –
Area = ‘ax b’ or ‘w x l’
‘Or’ Area of Triangle =
Where, a, b and c are the sides of triangle
And s = (a +b + c)/2
a) Area of Trapezium = ½
x (a + b) x h’
Where a and b are the parallel sides and h is
the perpendicular distance between them.
Step – 02
Ø The Area along the boundaries is
calculated as follows –
In figure,
O1, O2
= Ordinates
X1, X2 =
Chainages
Then
the area of the shaded portion =
In
the same way, the area between all pairs of ordinates are calculated and added
to obtain the total boundary area.
Therefore, the Total Area of the
field =
Area of
geometrical figure (ABCD) + Boundary
area (ABFEA)
(Step – 01)
+ (Step – 02)
·
Example–
A page of the field book of a cross-staff survey is given as follows.
Plot the required figure and calculate the relevant area.
·
Solution –
The figure can be plotted as follows –
The area is calculated as follows –
·
Area of the Field = 6455 sq.m
§ COMPUTATION OF AREA FROM PLOTTED PLAN:
The area can be calculated from the
following two ways –
Case 01 – Considering the Entire Area
The entire area is divided into regions of convenient geometrical shape
and the area is calculated as follows –
1)
By Dividing the Area into Triangles –
· The triangles are drawn so as to
equalize the irregular boundary line (as shown in figure).
· Then the base and altitude of the
triangles are determined according to the scale to which the plane was drawn.
· The area of the triangles are
calculated from area = ½ x base x altitude.
· The areas of the no of divided
triangles added to obtain the total area.
1)
By Dividing the Area into Squares–
· By this method squares of equal are
lined out on a piece of tracing paper.
· Each of the squares represents a
unit area which could be 1 Sq. cm or 1 Sq. m.
· Then the tracing paper is placed
over the plan and the number of full squares is counted.
· The total area is then calculated by
multiplying the number of squares by their unit area of each squares.
1)
By Drawing Parallel Lines and
Converting them to Rectangles –
· By this method, a series of equidistant
parallel lines are drawn on a tracing paper.
· The constant distance represents a
meter or centimetre.
· The tracing paper is placed over the
plan in such a way that the area is enclosed between the two parallel lines at
the top and bottom.
· Thus, the area is divided into a
number of stripes.
· The curved ends of the stripes are
replaced by perpendicular lines by give and take principles and a number of
rectangles are formed.
· The sum of the lengths of the
rectangles is then calculated –
The required area = ∑ Length of Rectangles x Constant
Distance
Case 02
1. In this method, a large
square or rectangle is formed within the area in the plan. Then ordinates are
drawn at regular intervals from the sides of the square to the curved boundary.
The middle area is calculated in the usual way. The boundary area is calculated
according to one of the following rules –
2.
The
Average Ordinate Rule
3.
The
Trapezoidal Rule
4.
Simpson’s
Rule
The Mid-Ordinate Rule –
In the
above figure –
·
O1 , O2 , O3 ........ On = Are
Ordinates at equal intervals.
·
l =
length of Base Line.
·
d =
Common Distance between ordinates.
·
h1 , h2 , h3, ........, hn = Are
Mid-Ordinates.
Then the Area of the Plot = h1 x d + h2 x d +
........ + hn x d
= d (h1 + h2
+ ........ + hn)
AREA = Common Distance x Sum of
mid-ordinates
1.
The Average Ordinate Rule –
In the above figure –
·
O1 , O2 , O3
........ On = Are Ordinates or Offsets at regular
intervals.
·
l =
length of Base Line.
·
n =
Number of divisions
(n+l) = Number of ordinates.
·
h1 , h2 , h3 ,
........ hn = Are Mid-Ordinates.
Then the Area of the Plot =
1.
The Trapezoidal Rule –
During the application
of The Trapezoidal Rule boundaries between the ends of the ordinates are
assumed to be straight. Thus the areas enclosed between the base line and the
irregular boundary line are considered as trapezoids.
In the above figure –
· O1, O2 ........ On = Are Ordinates or Offsets at regular
intervals. d = Common
distance
Thus the
Trapezoidal Rule may be started as follows –
·
To the sum of the
first and the last ordinate, twice the sum of the intermediate ordinates is
added.
·
This total sum is
multiplied by the common distance.
·
Half of this product
is the required area.
There are no such limitations for this rule and can be applied for any
number of ordinates.
1.
Simpson’s Rule –
In this rule, the boundaries between the ends of
ordinates are assumed to form an arc of a parabola. Hence Simpson’s Rule is
sometimes called the parabolic rule.
In the above figure –
· O1, O2, O3 = Are Ordinates or Offsets at regular
intervals.
· d = Common distance between the ordinates
Area
AFeDC = Area of trapezium AFDC + Area of segment FeDEF
·
Thus, the
rule stated as follows –
Ø The sum of the first and the last ordinate, four times
the sum of the even ordinates and twice the sum of the remaining odd ordinates
are added.
Ø This total sum is then multiplied by the common
distance.
Ø One third of this product is the required area.
This rule is applicable only when the number division
is even i.e. the number of ordinates is odd.
Example –
The
following offsets were taken from a chain line to an irregular boundary line at
an interval of 10 m –
0, 3.50,
4.50, 6.00, 4.20, 2.90, 0 m
Compute the
area between the chain line the irregular boundary line and the end off set by
–
A. The
Mid-ordinate Rule
B. The
Average-Ordinate Rule
C. The
Trapezoidal Rule
D. Simpson’s
Rule
Solution –
A. By
Mid-Ordinate Rule –
The mid-ordinates are –
A. By
Average-Ordinate Rule –
Here, d = 10 m & n = 6 nos (No of divisions)
Base Length = 10 X 6 = 60 m
Number of Ordinates = 7
A. By
Trapezoidal Rule –
Here, d = 10 m
A. By Simpson’s Rule –
Here, d = 10 m
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