area volume mine surveying

1. Lecture 17
Module 5: Area and Volume
Calculation

2. Computation of Area
 One of the main purposes of the topographical survey is to
determine the area of a land. 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.
 It is often need to know the areas of cross-section profiles to
determine the amount of earthwork need to do in mines.

3. Methods of Computation of Area
Area Calculation
Graphical Method Instrumental Method
From Field Data From Plotted Plan
Entire Area Boundary Area
Mid-Ordinate Average Ordinate Trapezoidal Simpson’s Rule

4. Computation of Area from field notes
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)
Where a, b, and c are the sides,
𝑠 =
𝑎+𝑏+𝑐
2

5. Area of the triangle = 1
2 × 𝑏 × 𝑕
Where, b = base, h = Altitude
Area of square =𝑎2
Where, a is the side of the square
Area of trapezium = 1
2 × (𝑎 + 𝑏) × 𝑑
Where, a and b are parallel sides, and d is the
perpendicular distance between them.

6.  The Area along the boundaries is calculated as follows:
Area of shaded portion =
𝑂1+𝑂2+(𝑋1+𝑋2)
2
𝑂1, 𝑂2 = 𝑂𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
𝑋1, 𝑋2 = 𝐶𝑕𝑎𝑖𝑛𝑎𝑔𝑒𝑠

7. From 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 equalize the irregular boundary line.
• Then the bases and altitude of the triangle are determined according to
the scale to which the plan was drawn. After this, the area of these
triangle's are calculated
Area =𝟏
𝟐 × 𝒃 × 𝒉
The area are then added to obtain the total area.

8. (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 counted. The total area is then
calculated by multiplying the number of squares by
the unit areas of each square.

9. 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

10.  The Mid-Ordinate Rule
Let, 𝑂1, 𝑂2, 𝑂3, … 𝑂𝑛 = 𝑂𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑒𝑞𝑢𝑎𝑙 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
l= length of base line.
d= common distance between ordinates,
𝑕1, 𝑕2, … . 𝑕𝑛 = 𝑚𝑖𝑑 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
Area of Plot = 𝒉𝟏 × 𝒅+𝒉𝟐 × 𝒅 + ⋯ + 𝒉𝒏 × 𝒅
= d(𝒉𝟏+𝒉𝟐 + ⋯ + 𝒉𝒏)
i.e., Area = common distance × sum of Mid-Ordinates.

11.  The Average Ordinate Rule
Let, 𝑂1, 𝑂2, 𝑂3, … 𝑂𝑛 = 𝑂𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
l= length of base line.
n= number of divisions,
n+1 =number of 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
Area =
𝑶𝟏+𝑶𝟐+⋯+𝑶𝒏
𝑶𝒏+𝟏
×l
i.e., Area =
𝑺𝒖𝒎 𝒐𝒇 𝑶𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔
𝑵𝒐 𝒐𝒇 𝑶𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔
× 𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐛𝐚𝐬𝐞 𝐥𝐢𝐧𝐞

12.  The Trapezoidal Rule
While applying the trapezoidal rule, boundaries between the ends
of the ordinates and 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, … 𝑂𝑛𝑎𝑟𝑒 𝑂𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑒𝑞𝑢𝑎𝑙 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
d= common distance.
1st Area =
𝑂1+𝑂2
2
×d
2nd Area =
𝑂2+𝑂3
2
×d
3rd Area =
𝑂3+𝑂4
2
×d
Last Area =
𝑂𝑛−1+𝑂𝑛
2
×d
Total Area =
𝑑(𝑂1+2𝑂1+2𝑂2+⋯2𝑂𝑛−1+𝑂𝑛)
2
=
𝐶𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
2
× [1st ordinate + Last ordinate + 2( sum of other ordinate)]

13.  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.
Let, 𝑂1, 𝑂2, 𝑂3𝑎𝑟𝑒 𝑡𝑕𝑟𝑒𝑒 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑂𝑟𝑑𝑖𝑛𝑎𝑡𝑒
d = common distance the ordinates
Area = AFeDC =Area of trapezium AFDC + area of Segment FeDEF
Here,
Area of Trapezium =
𝑂1+𝑂2
2
∗ 2𝑑
Area of Segment =
2
3
∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 =
2
3
∗ 𝐸𝑒 ∗ 2𝑑
=
2
3
*(𝑂2 −
𝑂1+𝑂3
2
) ∗ 2𝑑

14. So the area between first two division is given by
∆1=
𝑂1+𝑂2
2
∗ 2𝑑 +
2
3
∗(𝑂2 −
𝑂1+𝑂3
2
) ∗ 2𝑑
∆1=
𝑂1+4𝑂2+𝑂3
3
∗ 𝑑
Similarly, the area between next two division is given by
∆2=
𝑂3+4𝑂4+𝑂5
3
∗ 𝑑
Total Area =
𝑑(𝑂1+𝑂𝑛+4(𝑂2+𝑂4+⋯2(𝑂3+𝑂5+⋯ )
3
=
𝐶𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
3
∗ (1st ordinate + Last ordinate) +
4(sum of even ordinates)+2(sum of remaining odd ordinates)
Limitation of the Method
To apply this rule, the number of ordinates must be odd.
That is, the number of divisions must be even.

15. The Trapezoidal rule and Simpson rule
Trapezoidal rule
• The boundary between the
ordinates is considered to
be straight.
• There is no limitation. It can
be applied for any number
of ordinates.
• It gives an approximate
result.
Simpson rule
• The boundary between the
ordinates is considered to
be an arc of a parabola.
• To apply this rule, the
number of ordinates must
be odd. That is, the number
of divisions must be even.
• It gives a more accurate
result.

16. Problem 1
The following offset were taken from a chain line to an
irregular boundary line at an interval of 10m;
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

17. • By mid-ordinate rule: The mid-
ordinate are
𝐻1 =
0+2.5
2
= 1.25m
𝐻2 =
2.5+3.5
2
= 3.00m
𝐻3 =
3.5+5.00
2
= 4.25m
𝐻4 =
5+4.6
2
= 4.8m
𝐻5 =
4.6+3.2
2
= 3.9m
𝐻6 =
3.2+0
2
= 1.6m
Required Area
= 10(1.25+3.+4.25+4.8+3.9+1.6)
=10x 18.8 = 188 𝑚2
• By Trapezoidal Rule
Here d=10
Required Area
= 10/2 (0+0+2(2.5+3.5+5+4.6+3.2))
=5x37.60 =188 𝑚2
• By Simpson’s Rule
Here d=10
Required Area
=10/3 x(0+0+4(2.5+5+3.2)+2(3.5+4.6)
= 10/3 x (59.00)
= 196.66 𝑚2
• By average -ordinate rule:
Here d=10m and n=6 (no of div)
Base length = 10x6 =60m
Number of ordinates =7
Required Area = (1/7)x(60x(0+2.5+3.5+5+4.6+3.2+0))
= 60x 18.80 =161.14 𝑚2

18. Examples
The following offset were taken at 15 m interval
from a survey line to an irregular boundary line:
• 3.5,4.3,6.75,5.25,7.5,8.8,7.9,6.4,4.4,3.25m
• 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

19. By Trapezoidal Rule
Required Area
=
(15/2)x(3.5+3.25+2(4.3+6.75+5
.25+7.5+8.8+7.9+6.4+4.4)
=820.125 𝑚2
By 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
𝑜1to 𝑜9 and the area between 𝑜9
to 𝑜10 is found out by the
trapezoidal rule
• 𝐴1 =(15/2)x(3.5+4.4+4(4.3+5.25+
8.8+6.4)+2(6.75+7.5+7.9))
=756 𝑚2
• 𝐴2 =(15/2)x(4.4+3.25)
=57.38 𝑚2
Total Area = 𝐴1+𝐴2 = 813.38 𝑚2

20. 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.

21. 𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑕𝑒 𝑠𝑜𝑙𝑖𝑑 𝑙𝑖𝑛𝑒,
∴ 𝑃 = 𝑦0𝑥1 + 𝑦1𝑥2 + ⋯ . 0.0
𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑕𝑒 𝑑𝑜𝑡𝑡𝑒𝑑 𝑙𝑖𝑛𝑒
∴ 𝑄 = 0. 𝑦1 + 𝑥1𝑦2 + ⋯ . 0. 𝑦0
Required Area = ½ ( 𝑃- 𝑄)

22. Examples
• The following perperndicular offfset 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

23. 𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑕𝑒 𝑠𝑜𝑙𝑖𝑑 𝑙𝑖𝑛𝑒,
∴ 𝑃 = 5.25 x 5.5 + 6.5 x 12.7 +4.75x 25.5 + 5.2 x 40.5 +4.2 x 40.5 +0 x 0 + 0 x 0
=28.88+82.55+125.13+210.6+170.1
= 613.26 𝑚2
𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑕𝑒 𝑑𝑜𝑡𝑡𝑒𝑑 𝑙𝑖𝑛𝑒
∴ 𝑄 = 0 x 6.5 +5.5 x 4.75 +4.75+12.7x5.2+25.5x4.2+40.5x0+40.5x0+0x5.25
= 199.27 𝑚2
Required Area = ½ ( 𝑃- 𝑄)
= 206.995𝑚2

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

25. • 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.
• The other arm F is know 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 point of the tracing arm with
respect to the Vernier of the index mark I.

26. 𝑨𝒓𝒆𝒂 𝑨 = 𝑴 𝑭𝑹 − 𝑰𝑹 ± 𝟏𝟎𝑵 + 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

27. Volume calculation
In many engineering projects, earthwork involve excavation, removal, and
dumping of earth, therefore it is required to make good estimate of volume of
earthwork.
Computation of Volume by Trapezoidal and
prismoidal formula

28. • Computing of areas and volumes is an important
part of the office work involved in surveying. For
computation of the volume of earthwork, the
sectional area of the cross-section which are
taken to the longitudinal section during profile
leveling are first calculated.
• After calculating the cross sectional areas, the
volume of earth work is calculated by
• The Trapezoidal Rule
• The Prismoidal rule

29. Trapezoidal Rule (Average End Area Rule)
Volume (Cutting or Filling)
𝑉 =
𝑑(𝐴1 + 𝐴𝑛 + 2(𝐴2 + 𝐴3 + ⋯ + 𝐴𝑛−1)
2
i.e. Volume =
𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑛𝑎𝑐𝑒
2
x (first section area+ last section area
+2(sum of areas of other sections)
Prismoidal Formula
Volume,
𝑉 =
𝑑(𝐴1 + 𝐴𝑛 + 4 𝐴2 + 𝐴4 + 𝐴𝑛−1 + 2(𝐴3 + 𝐴5 + ⋯ + 𝐴𝑛−2)
3
i.e. Volume =
𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑛𝑎𝑐𝑒
3
x (first section area+ last section area
+4(sum of areas of even sections)+2(sum of areas of odd sections)

30. • The prismoidal formula is applicable when there are odd number of
sections. If the number of section are even, the end section is treated
separately and the area is calculated according to the trapezoidal rule. The
volume of the remaining section is calculated in usual manner by the
prismoidal formula. Then both the result are added to obtain the total
volume.
a) Prismoidal correction
• The difference between the volume obtained by the average end-area
method and the prismoidal method is referred to as the prismoidal
correction (𝐶𝑝). The correction formula is related to the distance (L)
between two end-areas. The center heights(h) of an earthwork section
(cut or fill) at the two end-areas. and the width (w) of an earthwork
section (from slope intercept to slope intercept) at the two end-areas.
𝐶𝑝 =
𝐿 × (𝑕1−𝑕2) × (𝑤1−𝑤2)
12

31. Example
• An embankment of width 10 m and side slope 1
1/2: 1 is required to be made on a ground which
is level in a direction traverse to Centre line. The
central height at 20 m intervals are as follows:
• 0.8,1.2,2.25,2.6,1.9,1.4 and 0.8
• Calculate the volume of earth work according to
(1) The trapezoidal formula
(2)The prismoidal formula

32. • Level section: Ground is level along the traverse direction
• Here, b=10m,s=1.5, interval =20m
• The cross sectional area are calculated by equation
• Area = (b + s h )h
• ∆1= 10 + 1.5 𝑋0.8 𝑋0.8 = 8.96 𝑚2
• ∆2= 10 + 1.5 𝑋1.2 𝑋1.2 = 14.16 𝑚2
• ∆3= 10 + 1.5 𝑋2.25 𝑋2.25 = 30.09 𝑚2
• ∆4= 10 + 1.5 𝑋2.6 𝑋2.6 = 36.14 𝑚2
• ∆5= 10 + 1.5 𝑋1.9 𝑋1.9 = 24.42 𝑚2
• ∆6= 10 + 1.5 𝑋1.4 𝑋1.4 = 16.94 𝑚2
• ∆7= 10 + 1.5 𝑋0.9 𝑋0.9 = 10.22 𝑚2
• Volume according to trapezoidal rule
• V= (20/2)( 8.96+10.22+2(14.16+30.09+36.14+24.42+16.94)
• = 10 (19.18+242.10)
• =2612.80 𝑚3

33. Volume according to prismoidal formula
V=
(20/3)(8.96+10.22+4(14.16+36.14+16.94)+2(30.
09+24.42))
=2647.73𝑚3
example
Q: calculate the volume of earthwork in an
embankment for which the cross-sectional areas
at 20 m interval are as follows
Distance 0 20 40 60 80 100 120
C/S area
(𝑚3)
38 62 74 18 22 28 13

34. Contour (m) 270 275 280 285 290
Area (m2)
2050 8400 16300 24600 31500
Example: The areas enclosed by the contours in the lake are as follows:
Calculate the volume of water between the contours 270 m and 290 m by:
i) Trapezoidal formula
ii) Prismoidal formula
Volume according to trapezoidal formula:
=5/2{2050+31500+2(8400+16300+24600)}
=330,250 m3

35. VOLUME OF A FRUSTRUM
VOLUME OF A PYRAMID
where A is the area of the base and h is
the height of the pyramid.

36. Q : water enters a sump having the shape of an
inverted frustum at a rate of 500𝑚3
/𝑕. the
sump is initially filled up to 2.0m height. The
time taken in days to fill the remaining part of
the sump (rounded off to one decimal places ) is