ERT252/4
GEOMATICS ENGINEERING
AREAS &
VOLUME
MRS SITI KAMARIAH MD SA’AT
LECTURER
SCHOOL OF BIOPROCESS ENGINEERING
sitikamariah@unimap.edu.my
AREAS


The computation of areas may be based on data
scaled from plans or drawings, or direct from the
survey field data.
Prevalent methods of areas determination.

Field measurements
 Dividing
the areas into geometrical figures, offsets from
base line and coordinates.

Plan measurements
 Based
on measurements from scaled plan or by the use of
planimeter.
Unit Conversion
Areas by geometrical figures

Trigonometric formula
c 2 sin 2 B
A
4
a (c  a )(c  a )
A
2
2
b cot B
A
2
b 2 tan A
A
2

Oblique triangle
ab
A
2
Areas from offset

Trapezoidal Rule
O1  O2
d
2
O  O3
Area of second trapezoid  2
d
2
O  On
Area of last trape zoid  n 1
d
2
O  On 
 O  O2 O2  O3
A  d 1

 ...  n 1

2
2
 2

Area of first trap ezoid 

Simpson’s Rules

Only for odd number of segments
d
A  O1  4O2  2O3  4O4  ...  2On  2  4On 1  On 
3
d
A  [(O1  On )  4(O2  O4  ...  On 1 )  2(O3  O5  ...On  2 )
3
Example 1




A tract of land has three straight boundaries AB, BC, and
CD. The fourth boundary DA is irregular. The measured
lengths are as under:
AB = 135 m, BC = 191 m, CD = 126 m, BD = 255 m.
The offsets measured outside the boundary DA to the
irregular boundary at a regular interval of 30 m from D,
are as below:
Determine the area of the tract.
Solution Example 1


Area ABD = 11604.42 m2
Area BCD = 11608.76 m2
Solution Example 1

A3 = 576 m2

So, total area = A1+A2+A3 = 23789.18 m2
Areas by coordinates

By coordinates
Y (N)
X (E)
YA
XA
YB
XB
YC
XC
YD
XD
YA
XA
1
Area  [(YA X B  YB X C  YC X D  YD X A )  ( X AYB  X BYC  X CYD  X DYA )]
2
1
Area  YA ( X B  X D )  YB ( X C  X A )  YC ( X D  X B )  YD ( X A  X C )
2
Example 2


The coordinates of traverse stations of a closed
traverse ABCDE are given in Table 8.1.
Calculate the area enclosed by the traverse.
Ans: 148300 m2
Earthwork Sections
Example 3

Calculate the area of cross-section that has breadth
of formation as 10m, center height as 3.2 m and
side slopes as 1 vertical to 2 horizontal.
Solution Example 3

Given
b = 10 m
 h = 3.2 m
 s= 2


Area = 52.48 m2
VOLUMES



Persons engaged in surveying are often called to
determine volumes of various types of material.
The most common unit of volume is cubic having
edges of unit length. Cubic feet, cubic yard or cubic
meter.
There are 3 methods to determining volume of land
for cut and fill works:
Cross section Method
 End Areas Method
 Prismoidal Method

Cross Section Method


The cross section method is employed almost
exclusively
for computing volumes on linear
construction projects such as highways, railroads, and
canals.
Cut slopes of 1:1 ( 1 horizontal to 1 vertical ) and fill
slope of 1-1/2:1 satisfactory for ordinary loam soil.
Earthwork Sections
Types Of Cross Sections


In flat terrain the level section (a) is suitable . The
three level section (b) is generally used where
ordinary ground level is prevail. Rough topography
may require a five level section (c).
More practically an irregular section (d). A transition
section (e), and slide hill section (f) occur when
passing from cut to fill and on slide –hill locations.
Types Of Cross Sections


The width of base b , the finished roadway is fixed
by project requirements and it is usually wider in
cuts than on fills to provide for drainage ditches.
Slide slops in fill are flatter than those in cuts where
soil remains in its natural state.
End Areas Method

A1 and A2 are end areas at two stations separated
by a horizontal distance L.
The volume between two stations then equal to :
Ve =
A1  A2
xL
2
Prismoidal Method

The prismoidal formula applies to volumes of all
geometric solids that can be considered prismoids.
Prismoidal Method
• The prismoidal formula gives very nearly correct
volume of earthwork even for irregular end sections
and sides that are warped surfaces.
• The prismoidal formula though being more accurate
than end-areas rule, in practice the end-areas rule is
more frequently adopted because of the ease of its
application.
• End-areas rule gives the computed volumes generally
too great which is in favour of contractor.
Prismoidal Method

The difference between the volumes obtained by the
average–end-areas formula and the prismoidal
formula is called the prismoidal correction Cp .
Cp 
L
c1  c2 w1  w2 
12
Where :
Cp is the volume of the prismoidal correction
c1 and c2 are center heights in cut or in fill
W1 and W2 are widths of sections from slope intercept to
slope intercept
*If the product ( c1 - c2 ) ( w1 – w2 ) is minus , the prismoidal
correction is added rather than subtracted from the end-area
volume.
Volume by Spot Level
where
 A = the area of the square or rectangle,
 ΣhI = the sum of the vertical depths common to one prism,
 Σh2 = the sum of the vertical depths common to two prism,
 Σh3 = the sum of the vertical depths common to three prism,
 Σh4 = the sum of the vertical depths common to four prism.
Volume from Contour



By cross-sections
By Equal Depth Contours
By horizontal planes
Sources of Error
1.
2.
3.
4.
Making errors in measuring field cross-sections, e.g.
Not being perpendicular to the centerline.
Making errors in measuring end areas.
Failing to use the prismoidal formula where it is
justifies.
Carrying out areas of cross sections beyond the
limit justified by the field data.
Exercise:

For the tabulated, calculate the volume of excavation
in cubic meter between stations 10+00 and 15+00.
Station
Cut End Area (m2 )
10 +00
24.0
11 +00
30.6
12 +00
43.5
13 +00
37.4
14 +00
22.7
15 +00
10
THANK YOU