Area & Volume

Syllabus
• Area of a irregular figure by Trapezoidal
rule, Mid-ordinate rule, average- ordinate
rule, Simpson’s rule, various coordinate
methods, planimeter.
• Computation of volume by trapezoidal and
prismoidal formula

Compution of Area
• The term area in the context of 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
• Area may be expressed in the following units
• Square metre
• Hectares
• Square feet
• Acres

Various methods of Computation of Area
Area
Graphical Method

From field Notes

Instrumental Method

From Plotted Plan

Entire Area
Mid Ordinate Avg Ordinate

Boundary Area
Trapezoidal Simpson’s Rule

Computation of Area from Field Notes
This is done in two steps
Step-I
• In cross-Staff survey, the area of field can be
directly calculated from the field notes. During
survey work the whole area is divided into
some
geometrical
figures,
such
as
triangle,
rectangles,
square,
and
trapeziums, and then the area is calculated as
follows
• Area of Triangle= √s (s-a) (s-b) (s-c)

Computation of Area from Field
Notes
Where a, b and c are the sides,
s = a+ b+ c
2
Or Area of the triangle = ½ x b x h
Where, b= base
h= Altitude
Area of square= a2
Where a is the side of the square
Area of trapezium= ½ (a+b) x d
Where a and b are the parallel sides, and d is the
perpendicular distance between them.

Notes
• The Area along the boundaries is calculated as
follows
• O1, O2= Ordinates
• X1, X2, = Chainages
• Area of Shaded Portion= O1 + O2 x (X1 + X2 )
2

Notes

Notes
• Similarly, the areas between all pairs of
ordinates are calculated and added to obtain
the total boundary area.
• Hence, Total area of the field = area of
geometrical figure + Boundary areas
(Step-1 + Step-2)
= Area of ABCD + Area of ABEFA

Computation of Area from Plotted
Plan
Case-I Considering the entire area
• The entire area is divided into regions of a convenient
shape, and calculated as follows:
(a) By dividing the area into triangles
• The triangle are so drawn as to equilize the irregular
boundary line.
• Then the bases and altitude of the triangles are determined
according to the scale to which the plan was drawn. After
this, the area of these triangles are calculated
• (area= ½ x base x altitude)
• The area are then added to obtain the total area.

Plan

Plan
(b) By dividing the area into Squares
• In this method, squares of equal sizes are ruled
out on a piece of tracing paper. Each square
represents a unit area, which could be 1 cm2
or 1 m2. The tracing paper is placed over the
plan and the number of full squares are
counted. The total area is then calculated by
multiplying the number of squares by the
unit area of each square.

Plan
(c) By drawing parallel lines and converting them to
rectangles
In this method, a series of equidistant parallel lines are
drawn on a tracing paper. The constant distance represents a
metre 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 strips. The curved end of the strip are
replaced by perpendicular lines and a number of rectangles
are formed. The sum of the length of the rectangles is then
calculated.
Then,
Required area= ∑ Length of Rectangle x constant distance

Plan
Case II
• 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 side of the square to the
curved boundary. The middle area is calculated
according to one of the following rules:
• The Mid-Ordinate rule
• The Average Ordinate Rule
• The Trapezoidal Rule
• Simpson’s Rule.

Plan
• The Mid-Ordinate Rule

Plan
• Let, O1, O2, O3….. On= Ordinates at equal
Intervals
• l= length of base line,
• d= common distance between ordinates,
• h1, h2, …hn= mid ordinates
• Area of Plot= h1 x d+ h2 x d + .. + hn x d
• = d(h1+ h2+ … hn)
• i.e. Area= Common Distance x Sum of MidOrdinates.

The Average-Ordinate Rule
• Let, O1, O2, O3….. On= Ordinates Regular
Intervals
• l= length of base line,
• n= number of divisions,
• n+1= number of ordinates
• Area= O1 + O2+ O3….. On x
l
On+1
• Area= Sum of Ordinates x length of base line
No of Ordinates

The Trapezoidal Rule
• While
applying
the
trapezoidal
rule, boundaries between the ends of the
ordinates are assumed to be straight. Thus the
area enclosed between the base line and the
irregular boundary lines are considered as
trapezoids

Let, O1, O2, O3….. On = Ordinate at equal intervals
d = common distance
1st Area= O1+ O2 x d
2
2nd Area= O2+ O3 x d
2
3 rd Area= O3+ O4 x d
2
Last Area= On-1+ On x d
2
Total Area = d [O1 + 2O1 +2O2….. 2 On-1 + On ]
2
= Common distance (1st ordinate + Last Ordinate+ 2 (sum of other Ordinate)
2

Thus, the trapezoidal rule may be stated as
follows:
• To the sum of the first and the last
ordinate, twice the sum of intermediate
ordinates is added. This total sum is
multiplied by the common distance, Half of
this product is the required area

Simpson’s Rule
• In this rule, the boundaries between the ends of
ordinate are assumed to form an arc of a
parabola. Hence Simpson’s rule is sometimes
called the parabolic rule.

Simpson’s Rule
Let,
O1, O2, O3 = Three Consecutive Ordinate
D= common distance between the ordinates
Area= AFeDC= Area of trapezium AFDC + area of Segment
FeDEF
Here,
Area of Trapezium = O1 + O2 x 2d
2
Area of Segment= 2 x area of parallelogram FfdD
3
= 2 x Ee x 2d = 2 x {O2- O1 + O3 } x 2 d
3
3
2

Simpson’s Rule
So, the area between the first two divisions,
∆ = O1 + O2 x
2d +
2 {O2- O1 + O3 } x 2 d
2
3
3
2
= d ( O1 + 4 O2 + O3)
3
Similarly, the area between next two divisions,
∆2
= d ( O3 + 4 O4 + O5)
3
Total Area= d ( O1 + On + 4 (O2 + O4+ …+ 2 (O3+ O5+…)
= Common distance ( 1st ordinate + last Ordinate)
3
+ 4 (sum of even ordinates)
+ 2 (sum of remaining odd Ordinates)

Simpson’s Rule
Thus, the rule may be stated as follow:
• To the sum of the first and the last ordinate, for
times the sum of even ordinates and twice the sum
of the remaining odd ordinates are added. This
total sum is multiplied by the common distance.
One-third product is the required area.

Limitation
• To apply this rule, the number of ordinates
must be odd. That is, the number of
divisions must be even.

The Trapezoidal rule and Simpson rule
Trapezoidal Rule

Simpson’s Rule

The boundary between the ordinates is The Boundary between the ordinates is
considered to be straight.
considered to be an arc of a parabola.

There is no limitation. It can be applied To apply this rule, the number of ordinates
for any number of ordinates.
must be odd. That is, the number of
divisions must be even.

It Gives an approximate Result.

It gives a more accurate result.

Examples
The following offset were taken from a chain line to an irregular
boundary line at an interval of 10 m;
• 0, 2.5, 3.5, 5.0, 4.6, 3.3, 0 m
• Compute the area between the chain line, the irregular
boundary line and the end offset by;
•
•
•
•

(a) The mid-Ordinate Rule
(b) The average- Ordinate Rule
(c) The trapezoidal rule
(d) Simpson’s Rule

Examples
By mid-ordinate rule: The mid- Ordinate are
H1= 0 + 2.5
= 1.25.
2
H2= 2.5 + 3.5 = 3.00
2
H3= 3.5 + 5.00 = 4.25 m
2
H4= 5.00+ 4.6 = 4.8 m
2
H5 = 4.6 + 3.2 = 3.9 m
2
H6= 3.2 + 0 = 1.6 m
2

Examples
Required Area
• = 10 (1.25 + 3.00 + 4.25 + 4.8 + 3.9 + 1.6)
• = 10 x 18.8 = 188 m2
• By average-ordinate rule:
• Here d= 10 m and n= 6 (no of div)
• Base length= 10 x 6 = 60 m
• Number of ordinates= 7
• Required Area= 60 x (0+ 2.5 +3.5+5+4.6+3.2+0)
•
7
•
= 60 x 18.80 = 161.14 m 2

Examples
By Trapezoidal Rule:
• Here d= 10
• Required Area
• = 10/2 ( 0 + 0 + 2 ( 2.5 + 3.5+ 5.0+ 4.6+ 3.2))
• = 5 x 37.60= 188 m2
• By Simpson’s Rule
• d= 10
• Required Area
• = 10 (0 + 0 + 4 (2.5+ 5.0+ 3.2) + 2 ( 3.5+ 4.6)
•
3
• 10 x 59.00
• 3
• 196.66 m2

Examples
• The following offset were taken at 15 m interval
from a survey line to an irregular boundary line:
• 3.5, 4.3, 6.75, 5.25, 7.5, 8.8, 7.9, 6.4, 4.4, 3.25 m
• Calculate the area enclosed between the survey
line, the irregular boundary line, and the first and
last offset by:
• (a) The Trapezoidal Rule
• (b) Simpson’s Rule

Examples
By Trapezoidal Rule

• Required Area
• = 15 (3.5 + 3.25 + 2 (4.3+ 6.75+ 5.25 + 7.5+ 8.8 + 7.9 + 6.4+ 4.4)
• 2
• = 15 (6.75 + 102.60) = 820.125 m2
• 2

Examples
Simpson’s Rule
• If this rule is to be applied, the number of ordinates must be odd but
here the number of ordinate is even (ten)
• So, Simpson’s rule is applied from O1 to O 9 and the area between
O9 and O 10 is found out by the trapezoidal rule.
• A1= 15 (3.5 + 4.4+4(4.3+5.25+8.8+6.4) + 2 (6.75 + 7.5 + 7.9)
•
3
• = 15 (7.9 + 99+ 44.3)= 756 m2
•
3
• A2= 15 (4.4 + 3.25) = 57.38 m2
•
2
• Total Area= A1 + A2 = 756 + 57.38 = 813.38 m2

Examples
• The perpendicular offsets taken at 10 m intervals from a
survey line to an irregular
• boundary are 2.18 m, 3.2 m, 4.26 m, 6.2 m, 4.8 m, 7.20
m, 8.8 m, 8.2 m and 5.2 m. Determine the area
• enclosed between the boundary, survey line, the first
and the last offsets by
• (i) Trapezoidal rule (ii) Simpson’s rule.
ANS

• Area = 463.5 m2
•
= 474.333 m2

Co-ordinate Method of Finding Area
• When Offset are taken at very irregular intervals, then
the application of the trapezoidal rule and Simpson’s
rule is very difficult. In such a case, the coordinate
method is the best.

Sum of Products along the Solid line,
• ∑ P = (y0x1 + y1 x2 + … 0.0)
Sum of Products, along the dotted Line
• ∑ Q = ( 0.y1 + x1 y2 + …+ 0. y0)
• Required Area= ½ (∑ P - ∑ Q)

Examples
• The following perpendicular offset were taken from a
chain line to a hedge
Chainage (m) 0 – 5.5- 12.7- 25.5- 40.5
Offset (m)
5.25- 6.5- 4.75-5.2-4.2
• Taking g as the origin, the coordinates are arranged
as follows:

Examples
• Sum of products along the solid line,
• ∑ P = (5.25 x 5.5 + 6.5 x 12.7 + 4.75 x 25.5 + 5.2 x
40.5 + 4.2 x 40.5 + 0x 0 + 0 x 0).
• = 28.88 + 82.55 + 121.13 +210.6+ 170.1= 613.26 m2
• Sum of Products along dotted line,
• ∑ Q= (0 x 6.5 + 5.5 x 4.75 + 12.7 x 5.2 + 25.5 x 4.2 +
40.5 x 0 + 40.5 x 0 + 0 x 5.25).
• = 26.13 +66.04 +107.10= 199.27 m2
• Required Area= ½ (∑ P - ∑ Q)
•
= ½ (613.26-199.27) = 206.995 m2

Instrument Method
• The Instrument used for computation of area
from a plotted map is the planimeter. The area
obtained by planimeter is more accurate than
obtained by the graphical method. There are
various types of planimeter in use. But the
Amsler Polar Planimeter is the most common
used now.

Instrument Method
• The Constructional Details of the planimeter
are Shown;
• It consists of two arms. The arm A is known
as the tracing arm. Its length can be adjusted
and it is graduated. The tracing arm carries a
tracing point D which is moved along the
boundary line of the area. There is an
adjustable Support E which always keep the
tracing point just clear of the surface.

Instrument Method
• The other arm F is known as the pole arm
or anchor arm, and carries a needle pointed
weight or fulcrum K at one end. The weight
forms the centre of rotation. The other end of
the pole arm can be pivoted at point P by a ball
and socket arrangement.
• There is a carriage B which can be set at
various points of the tracing arm with respect
to the vernier of the index mark I.

Instrument Method
• The carriage consists of a measuring wheel W
and a vernier V. The wheel is divided into 100
divisions and the vernier into 10 divisions. The
wheel and the vernier measures reading up to 3
places of decimal (i.e. 0.125, 0.174)
• The wheel is geared to a counting disc which is
graduated into 10 divisions for ten complete
revolutions of the wheel, the disc shows a
reading of one divisions

Instrument Method
• Thus the planimeter shows a reading of four
digits (i.e. 1.125, 1.174 etc.)
• The counting disc shows units
• The wheel shows tenth and hundredth and
vernier shows thousandths
• The planimeter rests on the tracing
point, anchor point and the periphery of the
wheel.

Procedure of finding area with a Planimeter
• Procedure of finding the area with a planimeter
• The vernier of the index mark is set to the exact
graduation marked on the tracing arm
corresponding to the scale as obtained from the
table.
• The anchor point is fixed firmly in the paper
outside or inside the figure. It should be
ensured that the tracing point is easily able to
reach every point on the boundary line.

Procedure of finding area with a
planimeter
• But it is always preferable to set the anchor
point outside the figure. If the area is very large.
It can be divided into a number of sections.
• A good starting point is marked on the
boundary line. A good starting point is one at
which the measuring wheel is dead. i.e. a point
where the wheel does not revolve even for a
small movement of the tracing point.

Planimeter
• By observing the disc, wheel and vernier, the
initial reading (IR) is recorded.
• The tracing point is moved gently in a clockwise
direction along the boundary of the area.
• The number of times the zero mark of the dial
passes the index mark in a clockwise or
anticlockwise direction should be observed.
• Finally, by observing the disc, wheel and vernier
the final reading (FR) is recorded.

planimeter
• Then, the area of the figure may be obtained from
the following expression.
• Area A= M (FR- IR 10 N + C)
• Where,
• M= Multiplier given in the table
• N= Number of times the zero mark of the dial
passes the index mark,
• C= the constant given in the table
• FR= Final Reading
• IR= Initial Reading.

Zero Circle in a Planimeter
• When the tracing point is moved along a circle without
rotation of the wheel (i.e. when the wheel slides
without any change in reading), the circle is known
as ‘ Zero Circle’ or ‘Circle of Correction’. The zero
circle is obtained by moving the tracing arm point in
such a way that the tracing point makes an angle of
90 0 with the anchor arm.
• The anchor point A is the centre of rotation and AT (R’)
is known as the radius of the zero circle.
• When the anchor point is inside the figure the area of
the zero circle is added to the area computed by
planimeter.

Finding Radius of Zero Circle
•
•
•
•
•
•
•
•
•
•
•

When the wheel is outside of the pivot point,
A= Anchor Point.
P= Pivot point
T= Tracing Point
W= Wheel
Let TP= L
PW= L1
AP= R
AT= R’ (Radius of Zero Circle)
From right-angled triangle AWT ( Angle AWT= 90 0)
AT 2 = AW 2 + TW 2

Or
• R’ 2= (AP2- PW2) + TW2
• = (R2-L12) + (L + L1)2
• = R2- L12+ L2 + 2 LL1 + L12
• = R2 + L2 + 2LL1
• R’= R2 + 2 LL1 + L2

When the wheel is inside the pivot,
Let,
TP= L
PW= L1
AP= R
AT= R1
Angle AWT= 90 0
From the right-angled triangle AWT,
• AT2= TW2 + AW2
• = (TP-PW) 2 + (AP2- PW2)
• R12= (L-L1)2 + ( R2- L12)
• = L2- 2 LL1 + L12 + R2 – L1 2
• L2- 2 L L1 + R2
• R’= R2 - 2 LL1 + L2

Finding Area of Zero Circle
By measuring radius of Zero Circle
• Area of Zero Circle= ∏ R’2
When the wheel is placed beyond the pivot point
• Area of Zero Circle= A= ∏ R’2
= ∏ (R2 + 2LL1 + L2 )
• When the Wheel is placed between the pivot and the tracing
point
• Area of Zero Circle,
• A = ∏ R’
•
= ∏ (R2 + 2LL1 + L2 )

Finding Area of Zero Circle
From multiplier and Constant
• Area of Zero Circle
• A= M x C
Where,
• M= multiplier value given in table
• C= Constant given in table
By Planimeter
• A geometrical figure is considered whose actual area is known.
After this, the area of the figure is computed by the Planimeter
• Then,
• Area of Zero Circle
• = Actual Area- Area Computed by Planimeter.

Examples
• The following reading were recorded by a
planimeter with the anchor point inside the figure:
•
•
•
•

IR= 9.377.
FR= 3.336
M= 100 cm2,
C= 23.521

• Calculate the area of the figure when it is
observed that the zero mark of the dial passed the
index mark once in the anticlockwise direction.

Examples
Solution
• IR= 9.377 M= 100 cm2
• FR= 3.336 C= 23.521.
• N= -1 (for anticlockwise rotation)
• From the expression, A= M (FR-IR 10 N + C).
• We get,
• A= 100 (3.336- 9.377- 10 x 1 + 23.521) = 748 cm2

Examples
• The area of an irregular figure was measured with a
planimeter having the anchor point outside the figure.
The initial and final reading were 4.855 and 8.754
respectively. The tracing arm was set to the natural
scale. The scale of map was 1 cm = 5 m. Find the area
of the figure.

Examples
Given
• IR= 4.855
• FR= 8.754
• M= 100 cm2
• N= 0
• Scale 1 cm= 5 m
• C= 0
• So,
• Area, A = M (FR- IR)
•
= 100 (8.754- 4.855)
•
= 389.9 cm2
• Required area= 389.9 x 25 = 9747.5 m2

Examples
• To determine the constants of a planimeter a 20 cm 20
cm area was measured with anchor point outside the plan
area. The zero mark of disc crossed the index in
clockwise direction once.
• The observed readings are
• Initial reading = 7.422
• Final reading = 1.422
• Determine the multiplying constant M.
• Solution: Area = M (F – I + 10 N)
• Now Area = 20 20 = 400 cm2
• F = 1.422, I = 7.422
• ∴ 400 = M (1.422 – 7.422 + 10 1)
• ∴ M = 100 cm2

Examples
A planimeter was used to measure the area of a map
once keeping anchor point outside the figure and
second time keeping it inside the figure. The
observations are as follows:
(i) When the anchor point was outside the figure:
• The zero of the dial did not pass the index at all.
(ii) When the anchor point was inside the map:

Examples
• Zero mark of the dial passed the fixed index
mark twice in anticlockwise direction.
• Find the (i) area of the map (ii) area of the
zero circle.
• Take multiplier constant M = 100 cm2.

Examples
• Solution:
(i) When the anchor point was outside the plan
• I = 1.486 F = 7.058, N = 0, M = 100 cm2.
• ∴ A = M (F – I + 10 N)
• = 100 (7.058 – 1.486 + 0)
• ∴ A = 557.2 cm2 Ans.
(ii) When the anchor point was inside the plan
• I = 3.486 F = 8.844 N = – 2, M = 100 cm 2.
• A = M (F – I + 10 N + C)
• 557.2 = 100 (8.844 – 3.486 – 10 2 + C)
• ∴ C = 20.214

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