260 Basic Engineering Mathematics
16 400 m
3
= 16 400 × 10
6 cm
3
.
Since 1 litre = 1000 cm 3 ,
capacity of reservoir =
16400 × 10 6
1000
litres
= 16 400 000 = 16.4 × 10
6 litres
Now try the following Practice Exercise
Practice Exercise 111 Volumes of irregular
solids (answers on page 352)
1. The areas of equidistantly spaced sections of
the underwater form of a small boat are as follows:
1.76, 2.78, 3.10, 3.12, 2.61, 1.24 and 0.85 m 2 .
Determine the underwater volume if the
sections are 3 m apart.
2. To estimate the amount of earth to be
removed when constructing a cutting, the
cross-sectional area at intervals of 8 m were
estimated as follows:
0, 2.8, 3.7, 4.5, 4.1, 2.6 and 0 m 3 .
Estimate the volume of earth to be excavated.
3. The circumference of a 12 m long log of timber
of varying circular cross-section is measured
at intervals of 2 m along its length and the
results are as follows. Estimate the volume of
the timber in cubic metres.
Distance from
one end (m)
0
2
4
6
Circumference (m) 2.80 3.25 3.94 4.32
Distance from
one end (m)
8
10
12
Circumference (m) 5.16 5.82 6.36
28.3 Mean or average values of
waveforms
The mean or average value, y, of the waveform shown
in Figure 28.6 is given by
y =
area under curve
length of base, b
If the mid-ordinate rule is used to find the area under the
curve, then
y =
sum of mid-ordinates
number of mid-ordinates
=
y 1 + y 2 + y 3 + y 4 + y 5 + y 6 + y 7
7
for Figure 28.6
y 1 y 2 y 3 y 4 y 5 y 6 y 7
d d d d
b
d d d
y
Figure 28.6
For a sine wave, the mean or average value
(a) over one complete cycle is zero (see
Figure 28.7(a)),
(b) over half a cycle is 0.637 × maximum value or
2
π
× maximum value,
(c) of a full-wave rectified waveform (see
Figure 28.7(b)) is 0.637 × maximum value,
(d) of a half-wave rectified waveform (see
Figure 28.7(c)) is 0.318 × maximum value
or
1
π
× maximum value.
V
0
t
V m
V
0
(a)
(b)
t
V m
(c)
V
0
t
V m
Figure 28.7
16 400 m
3
= 16 400 × 10
6 cm
3
.
Since 1 litre = 1000 cm 3 ,
capacity of reservoir =
16400 × 10 6
1000
litres
= 16 400 000 = 16.4 × 10
6 litres
Now try the following Practice Exercise
Practice Exercise 111 Volumes of irregular
solids (answers on page 352)
1. The areas of equidistantly spaced sections of
the underwater form of a small boat are as follows:
1.76, 2.78, 3.10, 3.12, 2.61, 1.24 and 0.85 m 2 .
Determine the underwater volume if the
sections are 3 m apart.
2. To estimate the amount of earth to be
removed when constructing a cutting, the
cross-sectional area at intervals of 8 m were
estimated as follows:
0, 2.8, 3.7, 4.5, 4.1, 2.6 and 0 m 3 .
Estimate the volume of earth to be excavated.
3. The circumference of a 12 m long log of timber
of varying circular cross-section is measured
at intervals of 2 m along its length and the
results are as follows. Estimate the volume of
the timber in cubic metres.
Distance from
one end (m)
0
2
4
6
Circumference (m) 2.80 3.25 3.94 4.32
Distance from
one end (m)
8
10
12
Circumference (m) 5.16 5.82 6.36
28.3 Mean or average values of
waveforms
The mean or average value, y, of the waveform shown
in Figure 28.6 is given by
y =
area under curve
length of base, b
If the mid-ordinate rule is used to find the area under the
curve, then
y =
sum of mid-ordinates
number of mid-ordinates
=
y 1 + y 2 + y 3 + y 4 + y 5 + y 6 + y 7
7
for Figure 28.6
y 1 y 2 y 3 y 4 y 5 y 6 y 7
d d d d
b
d d d
y
Figure 28.6
For a sine wave, the mean or average value
(a) over one complete cycle is zero (see
Figure 28.7(a)),
(b) over half a cycle is 0.637 × maximum value or
2
π
× maximum value,
(c) of a full-wave rectified waveform (see
Figure 28.7(b)) is 0.637 × maximum value,
(d) of a half-wave rectified waveform (see
Figure 28.7(c)) is 0.318 × maximum value
or
1
π
× maximum value.
V
0
t
V m
V
0
(a)
(b)
t
V m
(c)
V
0
t
V m
Figure 28.7