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Estimating Missing Precipitation Data

 
 

Interpretation of Average precipitation

Reasons due to which interpretation of precipitation is carried out.

  1. Gauges site is changed due to any reason.
  2. Short duration of storm occurred during installation
  3. Short duration of storm is missed due to absence of observer.
  4. Breaks in the record of missing data
  5. Station surrounded by dense forests
  6. Station surrounded by dense population
  7. Alteration in the instruments
  8. Abnormal results from a specific station

Conditions

  • Interpretation can only be carried out if:
    1. Topographically stations are not much different
    2. Range in altitude
    3. Shoot duration of rainfall spread over whole of the area which is steady for month and rainfall is calculated manually.

Interpretation

In order to avoid erroneous conclusions it is important to give the proper interpretation to precipitation data which often cannot be accepted at face value. For example a mean annual precipitation value for a station may have a little meaning if the gauge site has been changed significantly during period for which the average is computed.

Estimating missing precipitation data

Many precipitation stations have short breaks in their records because of absence of the observers or instrumental failure. It is must to estimate this missing record. The following method is adjusted for estimating this missing rainfall data.

Let there be three stations as close as possible to and evenly spaced around the station with missing record data i.e. Station X.

The rainfall data for these three stations i.e. A, B, C on the day for which the data at station is missing is collected. If normal precipitation at each of index station is within 10% of that for the station with missing record a simple arithmetic precipitation provides the estimated amount.

Px = (Pa + Pb + Pc)/3

If the difference is more than 10% than the normal ratio method is used. By this method precipitation Px at station X is in which N is the normal annual precipitation. Multiple linear regression will yield an equation.

Px = a + baPa + baPb + bcPc

Where a = 0

Average precipitation is required in many hydrological problems. For this purpose arithmetic mean method, Theism Polygon method and Isohytal methods are used.

Double mass analysis:

It is the technique to check the consistency of precipitation data for a particular station and to correct it. The consistency of precipitation data may be checked because:

  1. Change in location of instruments station
  2. Exposure
  3. Change of instrument (gauge)
  4. Change of observational procedure.

In this method the accumulated annual or seasonal precipitation with correct accumulated value of mean precipitation for a group of surrounding stations is compared.

  • A graph between the accumulated annual/seasonal and current accumulated precipitation is drawn.
  • The point where the line shows, change of slope, change the line.
  • Change in meteorological factor will not change the slope of the line because all the stations are equally affected.
  • The ratio of last line and slope of first line is found and is multiplied with data, prior to change of.

Determination of Average precipitation over area

A number of gauges in a certain catchment area are used to find average depth of precipitation of the area, the 3 general methods used are:

  1. Arithmetic mean method
  2. Theissen's Polygon Method
  3. Isohyets Method

i. Arithmetic mean method:

This is the simplest method and gives good results when:

  • The catchment area is flat
  • Precipitation gauges are evenly distributed over uniform area
  • The individual gauges do not vary too much from mean.
  • The average precipitation is determined as:

ii. Theissen's Polygon Method:

In this method adjacent stations are joined by straight line, thus dividing the area into series of triangles. Perpendicular bisectors are erected on each of these lines, thus, forming a series of polygons, each containing one and only one rainfall station. It is assumed that the entire area within a polygon is nearer to the rainfall station that is included in the polygon than to any other rainfall station.

The rainfall recorded at the station is assigned to that polygon. If P is the mean rainfall on the basin and A is the basin area then:

P =

P1, P2,.........Pn represents rainfall values at respective stations whose surrounding polygons have the areas A1, A2, .....An

 
Let us know in the comments what you think about the concepts in this article!

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