CHAPTAR-2
MEASUREMENT
OF
PRECIPITATION
Ch. Karamat Ali (C.E)
Department of Civil Engineering
University of Lahore
PRECIPITATION AND ITS MEASUREMENTS
Precipitation

Precipitation is the general term used for all forms of moisture
emanating from the clouds and falling to the ground. This
following of moisture is termed as precipitation.

The essential requirements for precipitation to occur are;
1.
Some mechanism is required “to cool the air sufficiently” to
cause condensation and droplet growth.
2.
Condensation nuclei are also necessary for formation of
droplets. They are usually present in the atmosphere in
adequate quantities.
3.
High cooling is essential for significant amount of
precipitation. This is achieved by lifting of air. Thus a
“meteorological phenomenon of lifting of air masses is
essential to result precipitation”.
PRECIPITATION AND ITS MEASUREMENTS
Types of Precipitation

Precipitation is often classified according
responsible for lifting of air as under:i)
Cyclonic Precipitation
ii)
Convective Precipitation
to
the
factors
iii) Orographic Precipitation
i. Cyclonic Precipitation
Cyclonic precipitation results from “lifting of air masses
converging into low pressure area of cyclone”. It is further
classified as tropical and extra-tropical. Tropical Cyclone results
into heavy rain falls and floods, whereas in extra-tropical, the rain
fall is lower and of longer duration.
Continued…
PRECIPITATION AND ITS MEASUREMENTS
ii. Convective Precipitation

Convective precipitation is caused by natural rising of warmer
lighter air in colder, denser surroundings, at high altitude. The
difference in temperature may result from unequal heating at
the surface, unequal cooling at the top of the air layer, or
mechanical lifting when air is forced to pass over a denser
colder air masses.

Convective precipitation is spotty and its intensity may
vary form light showers to cloud bursts.
iii. Orographic Precipitation

Orographic precipitation is due to the lifting of warm moisture
laden air mass due to topographic barriers (such as
mountains).

Precipitation is heavier on wind word slops and lighter
on the leeward slop and the over all rainfall is general
low.
PRECIPITATION AND ITS MEASUREMENTS
Forms of Precipitation
Drizzle : When the size of water droplets is under 0.5 mm, and its
intensity is < 0.01 mm per hour.
Rain
:
When the size of the drops is more than 0.5 mm. The upper
size of water drop is generally 6.25 mm, as drops greater
than this tend to break up as they fall through the air.
Glaze :
When the drizzle or rain freezes as it comes in contact with
cold objects, it is known as glaze.
Snow
: It is precipitation in the form of ice crystal resulting from
sublimation of water vapors directly to ice.
Hail
: Hail is lumps or bulbs of ice over 5 mm diameter formed by
alternate freezing or melting as they are carried up and down
in highly turbulent air currents.
PRECIPITATION AND ITS MEASUREMENTS
Measurement of Rain Fall

Rainfall is the main source of water, used for irrigation. Therefore,
a knowledge of its amount, character, seasons or periods is
important for designing, improvement and maintenance of
irrigation works”.

The amount of precipitation is defined as “the depth of water
which falls on a level surface”, and is measured by raingauge. The following are the main types of Rain-gauges:1.
Non-Automatic Rain-Gauges: This is also known as nonrecording rain-gauge. Symon’s Rain-Gauge is the instrument
general used for rainfall measurements.
2.
Automatic Rain-Gauge: These are integrating type recording
rain-gauges and are of three types;
i) Weighing bucket rain-gauge
ii) Tipping bucket rain-gauge
iii) Float type rain-gauge
PRECIPITATION AND ITS MEASUREMENTS
Non-Recording Rain-Gauges (Symon’s Rain-Gauge)

It consists of cylindrical vessel 127 mm (or 5”) in diameter with a
base enlarged to 210 mm (or 8”) diameter. The top section is
connected to a funnel provide with circular brass rim exactly 127
mm (5 inch) internal diameter.

The funnel shank is inserted in the neck of a receiving bottle which
is 75 to 100 mm (3 to 4”) diameter. A receiving bottle of rain-gauge
has a capacity of about 75 to 100 mm of rainfall and as during a
heavy rainfall this quantity is frequently exceeded, the rain
should be measured 3 or 4 times in a day on day of heavy
rainfall lest the receiver fill should overflow.

A cylindrical graduated measuring device is furnished with each
instrument, which reads to 0.2 mm. the rainfall should be
estimated to the nearest 01. mm.

The rain collected in the bottle is measured through provided
graduated cylinder.
PRECIPITATION AND ITS MEASUREMENTS
Symon’s Rain-Gauge
PRECIPITATION AND ITS MEASUREMENTS
Automatic Rain Gauges
i) Weighing Bucket Type Rain-Gauge

Self recording gauges are used to determine rates of rainfall
over periods of time. The most common type of self-recording
rain gauge is the “weighing bucket type”, as shown in Figure.

The rainfall from the receiver (30 cm) flows to the bucket through
the funnel.

The weight of the bucket is recorded by the mechanism of a
pen, chart and a clock-work revolving drum.

This

This type has the advantage of measuring any type of
precipitation like snow as it is based on weight.
graph of
accumulated rainfall against the lapsed period i.e. (mass
curve of the rainfall).
mechanism
of
the
instrument
give
the
PRECIPITATION AND ITS MEASUREMENTS
Weighing Bucket Type Rain-Gauge (Figure)
PRECIPITATION AND ITS MEASUREMENTS
PRECIPITATION AND ITS MEASUREMENTS
Mass Curve
PRECIPITATION AND ITS MEASUREMENTS
Automatic Rain Gauges
ii) Tipping Bucket Type Rain-Gauge

The tipping bucket type rain-gauge consists of 30 cm diameter
sharp edge receiver as shown in the figure. At the end of the
receiver is provided a funnel and allied mechanism.

A pair of tipping buckets are pivoted under the funnel in such a
way that when one bucket receives 0.25 mm (0.01 inch) of
precipitation “it tips, discharging its contents into a reservoir
bringing the other bucket under the funnel”.

The tipping operation is fitted with an electronic recording device
to compute the intensity of rainfall over time.
PRECIPITATION AND ITS MEASUREMENTS
Tipping Bucket Type Rain-Gauge (Figure)
PRECIPITATION AND ITS MEASUREMENTS
Automatic Rain Gauges
iii) Float Type Automatic Rain-Gauge

The working of a float type rain-gauge is similar to the weighing
bucket type gauge. A funnel receives the rain water which is
collected in a rectangular container.

A float is provided at the bottom of the container. The float is
raised as the water level rises in the container, its movement is
transmitted and being recorded by a pen moving on a clock-work
recording drum.

When the water level in the container rises so that the float
touches the top, the siphon comes into operation, and releases
the water; thus all the water in the box is drained out.
PRECIPITATION AND ITS MEASUREMENTS
Float Type Automatic Rain-Gauge
PRECIPITATION AND ITS MEASUREMENTS
Float Type Automatic Rain-Gauge (Graph)
PRECIPITATION AND ITS MEASUREMENTS
Site Selection for Gauging Station
i.
The site where a rain-gauge is set up should be an open place.
ii.
The distance between the Rain-gauge and the nearest object
should be at least twice the height of the object. In no case
should it be nearer to the obstruction than 30 metres.
iii.
The gauge should be placed on the level ground not upon a slop.
iv.
In the hill area, if suitable level area is not available then it should
be placed on the top of the hill.
v.
Site should be best shielded from high wind.
vi.
A fence, if erected to protect the gauge from cattle etc. should not
be less than twice its height.
PRECIPITATION AND ITS MEASUREMENTS
Advantages and Disadvantages of Recording Rain-Gauges
Following are the “advantages of recording type rain-gauge” over the
non-recording type:

The rainfall is “recorded automatically” and therefore, there is no
necessity of any attendant.

The recording rain-gauge also gives the “intensity of rainfall at
any time” while the non-recording gauge gives the “total rainfall
in any particular interval of time”.

As no attendant is required such rain-gauge can be installed in
far-off places also.

Possibility of human error is obviated.
Disadvantages.

It is costly comparing to non-recording gauge.

Fault may develop in electrical or mechanical mechanism or
recording the rainfall.
PRECIPITATION AND ITS MEASUREMENTS
Sources of Errors in Recording the Measurements
i.
The most serious error is the “deficiency of measurements due
to wind”. Vertical acceleration of air forced upwards over a
gauge gives and upward acceleration to precipitation about to
enter the gauge and “results in deficient catch”.
ii.
Inclination of gauge may “cause lesser collection”. A 10%
inclination gives about 15% low catch.
iii.
Tipping of the buckets may be affected due to rusting or
accumulation of dirt at the pivot.
iv.
Mistakes in reading the scale of the gauge.
v.
Dents in the collector rim may change its receiving area.
vi.
Funnel and inside surface require about 2.5 mm of rain to get
moistened when the gauge is initially dry. This may amount to
the extent of 25 mm per year in some areas.
PRECIPITATION AND ITS MEASUREMENTS
Example
The chart fixed to an automatic float type Rain Gauge gives the
result as shown in the table below;
PRECIPITATION AND ITS MEASUREMENTS
Example
The chart fixed to an automatic float type Rain Gauge gives the result
as shown in the table below;
Find
1.
2.
3.
4.
5.
Hourly Precipitation
Daily Precipitation
Time when the pointer reverted
Period of no Precipitation
Maximum Intensity of Precipitation
PRECIPITATION AND ITS MEASUREMENT
Computation of Average Rainfall Over a Basin
Average Rainfall intensity is measured over a catchment by
following commonly used method;

Arithmetic Mean Method.

Thiessen polygon method

Isohyetal method

Weighted Length Method
Continued…. 23
PRECIPITATION AND ITS MEASUREMENT
Arithmetic Mean Method

This is the simplest method of computing the average rain fall
over a basin.

The average rainfall is obtained by “dividing the sum of depths
recorded at different rain gauge stations of the basin” by the
number of rain gauge stations.
Pm = (P1+P2+P3+…..+Pn) / n
= 1/n ∑
n
i=l
Pi
Merits and Demerits
1.
2.
3.
4.
5.
6.
It is one of the simplest methods.
It is used to determine the approximate rainfall.
If the rainfall is uniformly distributed over the whole catchment,
method gives better results.
If the number of rain gauge stations are more and variation of the
individual record is not far from the mean; method is taken to be
accurate.
In hilly terrains, this method can yield fairly satisfactory results
To install maximum number of rain gauge stations in the catchment,
24
for better results, is costly and not always possible.
• Example:
Rainfall of five Rain Gauges is shown below.
Calculate the Average Rainfall.
Stations
Precipitation in mm
1
2
3
4
5
6
15.6
20.4
13.8
10.5
17.1
22.3
Total = 99.7 mm
Average Precipitation
∑P = 99.7 mm
Pav = 99.7/6
= 16.62 mm Ans
PRECIPITATION AND ITS MEASUREMENT
Thiessen Polygon Method

This is the weighted mean method.

The rain fall is never “uniform” over the entire area of the
basin or catchments, “but-varies in intensity and duration
from place to place”.

The rain fall recorded by each rain gauge station is weighted
according to the area, it represents.

This method is more suitable under the following conditions.
i.
For areas of moderate size ranging from say 750 sq km
to 3000 sq km.
ii.
When the rain gauge stations are few compared to the
size of the basin.
Continued…. 26
PRECIPITATION AND ITS MEASUREMENT
Thiessen Polygon Method (Figure-I)
27
PRECIPITATION AND ITS MEASUREMENT
Thiessen Polygon Method
Procedure

Join the adjacent rain gauge stations A,B,C,D,E etc. of the area
dividing the entire area in a series of triangles as sown in figure.

Draw the perpendicular bisector of each of these lines, as shown
by firm lines in figure.

The area enclosed by these perpendicular bisectors is served by
respective rain gauges. Thus these perpendicular bisectors form
a series of polygons around the Rain-gauge stations and
containing one and only one Rain-gauge station in each polygon.

The entire area within any polygon is nearer to the Rain-gauge
station contained there in than any other.

Find the area of each polygon and multiply it with the rainfall of
the respective Rain-gauge.
Continued…. 28
PRECIPITATION AND ITS MEASUREMENT
Thiessen Polygon Method
The Calculations

If P1, P2, P3, P4, P5, and P6 are the rainfalls in station and
A1,A2,A3,A4,A5 and A6 are the areas respectively then average
rainfall of the catchment is;
Pav = (P1A1+P2A2+P3A3+P4A4+P5A5+P6A6+…+PnAn)
A1+A2+A3+A4+A5+A6 +…+An
P=
n
(∑ i=l
PiA1 )/ A
29
PRECIPITATION AND ITS MEASUREMENT
Merits and demerits
i.
Results are accurate than arithmetical mean method.
ii.
The greatest limitation of the method is its flexibility. A
new Thiessen polygon diagram is to be drawn every
time for the catchment if there is a change in the gauge
network.
iii.
This method simply assumes linear variation of
precipitation between stations and assigns each
segment of area to the nearest station.
iv.
If gauging stations are few compared to the size of area,
Thiessen polygon method should be used.
v.
Station weights remain constants when the same
number of stations are used.
vi.
If catchment area is large and rain gauging stations are
also quite large in number, it is then adoptable to
computer computation.
Continued…. 30
PRECIPITATION AND ITS MEASUREMENT
Thiessen Polygon Method
• Example:
The map of a river basin is shown in the figure 1. Rainfall observation of
the available rain gauge stations are noted on the map itself. Draw the
Network of Thiessen Polygon and find out the Average Rainfall.
• Solution:
The Thiessen Polygon are drawn as explained earlier and are shown in
Figure-I. The area of each polygon has measured by planimeter and is
shown in following table.
The Map of the area is drawn into a scale of (1 cm=320 m) and the results
are tabulated as below.
Stations
Area of Thiessen Polygon
(A) cm2
Precipitation
in cm (P)
Product
AXP
A
B
C
D
E
F
G
H
I
J
K
L
112.25
53.50
120.0
62.5
119.0
144.0
72.0
130.0
62.5
85.0
110.0
40.0
62.5
70.7
67.5
85.0
77.5
80.0
82.5
55.0
52.5
67.5
60.0
57.5
7020.0
3745.0
8100.0
5312.5
9922.5
11520.0
5940.0
6950.0
3281.25
5737.5
6600.0
2300.0
∑A=1110.75
• Average Precipitation:
=
∑AP=26428.75
∑AP
∑A
= 76428.75
1110.75
= 68.8 cm
Ans
PRECIPITATION AND ITS MEASUREMENT
 Isohyetal Method

“An isohyets is a line joining places of equal rainfall
intensities on the rain fall map of a basin”.

An Isohyetal map showing contours of “equal rainfall”
presents a more accurate picture of the rain fall distribution
over the basin.

This method is more suited under the following conditions.
i.
For hilly and rugged areas.
ii.
For large areas over 5000 square km.
iii.
For areas where the net-work of rain gauge stations with
in the storm area is sufficiently dense, Isohyetal method
gives more accurate distribution of rainfall.
Continued…. 33
PRECIPITATION AND ITS MEASUREMENT
Isohyetal Method (Figure-II)
34
PRECIPITATION AND ITS MEASUREMENT
Procedure




From the rain fall data prepare the isohyetal map as shown in
figure.
Measure the areas between two successive isohyets with the help
of a planmeter.
Multiply each area by the mean rainfall between the isohyets.
Calculate the average rainfall by the following equation
The Calculation

Let the isohyets represent the rainfall P1, P2,…… Pn and areas
between the successive isohyets A1, A2,…… Pn-1 Average
precipitation is then calculated as;
Pav = A1[P1+P2/2]+A2[P2+P3/2]+….+An-1[Pn-1+Pn/2]
A1+A2+A3+A4+….+An-1
P = A1[P1+P2/2]+A2[P2+P3/2]+….+An-1[Pn-1+Pn/2]
A
Continued….
35
PRECIPITATION AND ITS MEASUREMENT
Merits and demerits
i.
ii.
iii.
iv.
v.
The isohyetal method permits the use and interpretation
of all available data and is well adapted to display and
discussions.
If the analyst has the knowledge of orographic effect, he
can use it in constructing the isohyetal map. Also he
must have knowledge of storm morphology. Then the
final map prepared by the experienced analyst should
represent a more realistic precipitation pattern than just
obtained from gauged data.
In the hilly and rugged basin, this method is most
suitable.
In sufficient dense network of rainfall stations within
storm area, this method may give reasonable accurate
indication of rainfall distribution.
Overall impression may be concluded that the isohyetal
method is superior to the other two methods.
36
PRECIPITATION AND ITS MEASUREMENT
Isohyetal Method
• Example:
Isohyets of different rainfall are shown in Figure-II. The rainfall and
areas of adjacent Isohyets are also given in the figure. Find out the
Average Rainfall of basin.
Solution:
i) Draw the Isohyets on the basis of Rainfall intensity catchment.
ii) Find out the area through planimeter between the two Isohyets.
iii) Find out the average of two Isohyets of particular area.
iv) Then multiply the area with the mean Isohyets
v) Find out the accumulative area and the product of each area and
Isohyets.
vi) Then the calculate the mean rainfall as under;
S. No.
Values of isohyets
bounding the strip cm
Mean Value
of Rainfall
‘P’ (cm)
Area
‘A’
Product
∑P av X A
1
2
3
4
5
6
7
30 _ 40
40 _ 50
50 _ 60
60 _ 70
70 _ 80
80 _ 90
90 _ 100
35.0
45.0
55.0
65.0
75.0
85.0
95.0
32.0
162.0
155.0
92.0
228.0
120.0
65.0
1120.0
7290.0
8525.0
5980.0
17100.0
10200.0
5575.0
∑A=854
55700.0
Pav =
∑Pm
∑A
= 55790
854
= 65.33 cmAns
PRECIPITATION AND ITS MEASUREMENT
Example
Find the “Average Precipitation” for the following given data by all
the three methods;
a) Arithmetic Mean
b) Polygon Method
PRECIPITATION AND ITS MEASUREMENT
Example
c) Isohyetal Method
PRECIPITATION AND ITS MEASUREMENT
Interpolation and Adjustment of Missing Data

Some times Rainfall Data in 1 or 2 stations may be missed. Inconsistency of average rainfall may occur in catchment due to
various reasons.
 exposure of station may be changed with growth of trees and
buildings. In such situation interpolation in the estimation in the
average rainfall is required.
 This interpolation a done by following methods depending upon
causes of inconsistency.
Arithmetic Mean Method



In order to determine the missing data at a particular station,
estimate this data, at lest three stations closed to the station of
missing data. It is necessary that these three stations are evenly
distributed around the station under consideration.
If normal precipitation at each of these selected stations is with in
10% of that station with missing data then simple arithmetical mean
of precipitation of those three stations will give the value of missing
station.
If PA, PB and PC are precipitation of the nearby station and Px is the
estimated the precipitation of missing station then
41
P = P +P +P /3
PRECIPITATION AND ITS MEASUREMENT
Interpolation and Adjustment of Missing Data
Normal Ratio Method

This method is used for interpolation when the normal
annual precipitation of, say, three nearby stations A, B and C
NA, NB and NC, differs from that of the NX of station X with
missing data from previous record by more than 10%.

If PA, PB, PC are average storm precipitation of A, B and C in
the year when average precipitation at X, i.e., PX is missing,
then.
PX = 1/3 [NX/ NA] x PA + [NX/NB] x PB + NX/NC x PC
42
PRECIPITATION AND ITS MEASUREMENT
Example:

The normal annual rainfall at station A,B,C and D in a basin are
80.97, 67,59, 76.28 and 92.01 cm respectively. In the year 1975,
the station D was inoperative and station A,B and C recorded
annual precipitation of 91.11, 72,23 and 79.89 cm respectively.
Estimate the rain fall at station D in that year.

Solution:

As the annual rainfall values vary more than 10 %, the Normal
ratio method is adopted

PD = 92.01/3 X [91.11/80.79 + 72.23/67.59 + 79.89/76.28]

PD = 99.48 cm
43
PRECIPITATION AND ITS MEASUREMENT
Example
• In a catchment area, daily precipitation was observed by 11 rain gauge
stations. On 2nd August 2005, the observations indicated that one gauge was
out of order. The observation taken by 10 rain gauges are as follows
• Estimate the missing data at ‘H’
Station
A
B
C
D
E
F
G
H
I
J
K
Prpt.
21
23
19
20
23
24
19
?
21
22
18
Solution
• Since there is not much variation in the precipitation data, a simple
arithmetic average of the precipitation observed at the 10 remaining stations
was taken as under
Precipitation at H = 21+23+19+20+23+24+19+21+22+18 = 210 = 21mm
10
10
PRECIPITATION AND ITS MEASUREMENT
Example
• The average annual precipitation at five rain gauge stations in a
catchment is as follows.
Station
P
Q
R
S
T
Avg. PPt
2400
2332
2431
2207
2231
• However the precipitation at station P was not available for the year
1996 because the rain gauge was out of order. The precipitation
observed at the other stations in1996 was as follows.
• Evaluate the precipitation at station ‘P’ during 1996.
Station
P
Q
R
S
T
PPT
?
2113
2200
2028
2095
PRECIPITATION AND ITS MEASUREMENT
Solution
Precipitation at P (in mm)
= ¼ (2400X2113 + 2400 X 2200 + 2400 X 2028 + 2400 X 2095)
2332
2431
= ¼ (2174+2172+2205+2253)
= 2201 mm
2207
2231
PRECIPITATION AND ITS MEASUREMENT
Interpolation and Adjustment of Missing Data
Double Mass Curve

A double mass curve shown in figure is drawn for a period of 29
years of a catchment to check the inconsistency of rainfall
record of a station (say, X) and accordingly to adjust the
incorrect results.

The cause of inconstancy may be;
i. Shifting of rainfall station to a new position
ii. Error in observation from a certain year.
iii. Change in the vicinity of the station due to growth of trees,
building, fencing, cutting of forest nearby changing the wind
pattern, etc.
iv. Rain gauge may be faulty from a certain period.
v. Site of instrument may be changed or replaced without
record.
Continued…. 47
PRECIPITATION AND ITS MEASUREMENT
Procedure

To draw this curve, a group of stations (say, 10) is taken as base
station in the neighborhood of the problem station “X”.

The accumulated rainfall of station X, (∑PX) and accumulated values
of average of group of base stations ∑Pav are calculated starting
from latest record.

The values ∑PX are plot as ordinate ∑Pa as abscissa for available
date of rainfall for the entire period.

In this plot a break in the slope will be been seen from the particular
year. Which indicates the years of change in precipitation regime of
station X.

The values at X beyond the period of change of regime the particular
year are corrected as under;

The original slope of the mass curve of rainfall upto the change of
years is extended by a dotted line as shown in the figure-III.
48
PRECIPITATION AND ITS MEASUREMENT

The correction to the data of rainfall at X from 1982 to 1971 is the
slope of the dotted line divided by the slope of the second mass
curve of 1983 to 1971, i.e. (c/b) / (s/b) = c/s
Thus the corrected data at X is;

Pcx = (Px) x c/s
All the inconsistent data of X from 1982 to 1979 are corrected by
multiplying by c/s, the value of which is obtained from the plot
measuring c and s as per scale
49
PRECIPITATION AND ITS MEASUREMENT
Double Mass Curve (Figure-III)
50
ANNUAL RAINFALL DATA FOR STATION M AS WELL AS THE
AVERAGE ANNUAL RAINFALL VALUES
Example
Annual rainfall data for station M as well as the average annual rainfall
values for a group of ten neighboring stations located in a
meteorologically homogeneous region are given below.
Test the consistency of the annual rainfall data of station M and correct
the record if there is any discrepancy. Estimate the mean annual
precipitation at station M.
Year
Annual Rainfall
of Station M
(mm)
Average Annual
Rainfall of the
group (mm)
Year
Annual
Rainfall of
Station M
(mm)
Average Annual
Rainfall of the
group (mm)
1950
676
780
1965
1244
1400
1951
578
660
1966
999
1140
1952
95
110
1967
573
650
1953
462
520
1968
596
646
1954
472
540
1969
375
350
1955
699
800
1970
635
590
Continued….
ANNUAL RAINFALL DATA FOR STATION M AS WELL AS THE
AVERAGE ANNUAL RAINFALL VALUES
Year
Annual Rainfall
of Station M
(mm)
Average Annual
Rainfall of the
group (mm)
Year
Annual
Rainfall of
Station M
(mm)
Average Annual
Rainfall of the
group (mm)
1956
479
540
1971
497
490
1957
431
490
1972
386
400
1958
493
560
1973
438
390
1959
503
575
1974
568
570
1960
415
480
1975
356
377
1961
531
600
1976
685
653
1962
504
580
1977
825
787
1963
828
950
1978
426
410
1964
679
770
1979
612
588
CALCULATION OF DOUBLE MASS CURVE
YEAR
Pm
(mm)
∑Pm (mm)
Pav
(mm)
Pav
(mm)
Adjusted values of Pm
(mm)
Finalized values of
Pm (mm)
1979
612
612
588
588
612
1978
426
1038
410
998
426
1977
825
1863
787
1785
825
1976
685
2548
653
2438
685
1975
356
2904
377
2815
356
1974
568
3472
570
3385
568
1973
438
3910
390
3775
438
1972
386
4296
400
4175
386
1971
497
4793
490
4665
497
1970
635
5428
590
5255
635
1969
375
5803
350
5605
375
1968
596
6399
646
6251
689.92
699
1967
573
6972
650
6901
671.95
672
1966
999
7971
1140
8041
1171.51
1172
1965
1244
9215
1400
9441
1458.82
1459
1964
679
9894
770
10211
796.25
796
Continued….
CALCULATION OF DOUBLE MASS CURVE
YEAR
Pm
(mm)
∑Pm (mm)
Pav
(mm)
Pav
(mm)
Adjusted values of Pm
(mm)
Finalized values of
Pm (mm)
1963
828
10722
950
11161
970.98
971
1962
504
11226
5801
11741
591.03
591
1961
531
11757
600
12341
622.70
623
1960
415
12172
480
12821
486.66
487
1959
503
12675
575
13396
589.86
590
1958
493
13168
560
13956
578.13
578
1957
431
13599
490
14446
505.43
505
1956
479
14078
540
14986
561.72
562
1955
699
14777
800
15786
819.71
820
1954
472
15249
540
16326
553.51
554
1953
462
15711
520
16846
541.78
542
1952
95
15806
110
16956
111.41
111
1951
578
16384
660
17616
677.81
678
1950
676
17060
780
18396
792.73
793
Total of Pm =
19004 mm
Mean of Pm=
633.5 mm
CALCULATION OF DOUBLE MASS CURVE