2. Problem 2: 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
4. By Simpson’s Rule
This rule is applicable odd number of ordinates 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 =
𝑑[(𝑂1:𝑂9):4(𝑂2:𝑂4:𝑂6:𝑂8):2(𝑂3:𝑂5:𝑂7)]
3
=
15 [ 3.5:4.4 :4 4.3:5.25:8.8:6.4):2(6.75:7.5:7.9 ]
2
= 756 𝑚2
• A2 =
𝑑(𝑂9:𝑂10)
2
=
15∗(4.4:3.25)
2
= 57.38 𝑚2
Total Area = 𝐴1+𝐴2 = 813.38 𝑚2
O1 O2 O3 O4 O5 O6 O7 O8 O9 O10
d
O1 O2 O3 O4 O5 O6 O7 O8 O9 O10
d
5. 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.
6. 𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑒 𝑠𝑜𝑙𝑖𝑑 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝑃 = 𝑦0𝑥1 + 𝑦1𝑥2 + ⋯ . 0.0
𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑒 𝑑𝑜𝑡𝑡𝑒𝑑 𝑙𝑖𝑛𝑒 is given by
𝑄 = 0. 𝑦1 + 𝑥1𝑦2 + ⋯ . 0. 𝑦0
Required Area = ½ ( 𝑃- 𝑄)
7. Problem 3: 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:
8. 𝑆𝑢𝑚 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑙𝑜𝑛𝑔 𝑡𝑒 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 𝑡𝑒 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
9. 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
10. • 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.
11. 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
12. 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
13. • 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
14. 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)
15. • 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
16. 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
17. • 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
18. 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
19. 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
20. VOLUME OF A FRUSTRUM
VOLUME OF A PYRAMID
where A is the area of the base and h is
the height of the pyramid.
21. 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
