The area may be calculated by the rule:
· ü Mid-ordinate Rule
· ü Average ordinate Rule
· ü Trapezoidal rule
· ü Simpson’s Rule
1.Mid-ordinate rule:
Area = (O1+O2+……….+on)d
O1, O2 = the ordinates at the mid points of each division
n = number of divisions
L = Length of base line= nd
d = Distance of each division
2.Average ordinate Rule
Area = ((O1+O2+……On)/(n+1))L
O1, O2….= Ordinates at the end of each division
3. Trapezoidal Rule
Area= (((O0+On)/2)+(O0+O2+……On-1))d
4. Simpson’s Rule
Area = ((d/3) x ((O0+On+4(O1+O3+On-1)+2(O2+O4+….+On-2))
Example: 1
The following perpendicular offsets were taken at 10 meter intervals from a survey line to an irregular boundary line.
3.15m, 4.3m, 8.2m, 5.6m, 6.85m, 7.6m, 4.2m, 5.6m, 4.3m
Calculate the area enclosed between the survey line, the irregular boundary line, and first and last offsets, by the application of
a) Average ordinate rule
b) Trapezoidal rule
c) Simpson’s rule
Average
d = the interval between the offset = 10 m
n = number of divisions = 8
n+1 = number of ordinates = 8+1 =9
L= Length of the base line = 8 x 10 = 80m
a) Average ordinate rule
Area = ((O1+O2+……On)/(n+1))L
O1, O2….= Ordinates at the end of each division
Area = (( 3.15 + 4.3 + 8.2+ 5.6 + 6.85 + 7.6 + 4.2 + 5.6 + 4.3 )/(8+1)) x 80 =442.66m2
b) Trapezoidal rule
Area= (((O0+On)/2)+(O0+O2+……On-1))d
Area = (3.725 + 42.35) x10 =460.75m2
c) Simpson’s rule
Area= 10/3(7.45 +92.4 +38.5) = 461.167m2
