[gradsusr] need help regarding time conversion

[Eric Altshuler]

First you need to find the height of a column of water of area 1 m^2 and mass 1 kg. Since the density of liquid water is ~1000 kg/m^3, the height of the column must be 0.001 m, or 1 mm. Numerically, the "conversion" from kg/m^2 to mm of liquid water is 1. To convert the precip rate to mm/day, multiply by 86400. To get it in mm/month, multiply by the number of days in the month you're interested in. 

----- Original Message -----
From: "asmat ullah" <asmat-786 at hotmail.com> 
To: gradsusr at gradsusr.org 
Sent: Tuesday, November 26, 2013 4:12:02 AM 
Subject: [gradsusr] need help regarding time conversion 


pr_1_transformed_regrid, Season_1_per_year
             VARIABLE : Average Precipitation [ x=60.75:74.3 at Average y=23.3:30.25 at Average] (kg m-2 s-1) every Dec ((kg m-2 s-1))
             SUBSET   : 22 points (TIME)
             CALENDAR : NOLEAP
 16-DEC-1979 /  1:  2.985E-07
 16-DEC-1980 /  2:  1.188E-06
 16-DEC-1981 /  3:  3.238E-06
 16-DEC-1982 /  4:  7.722E-07 i have these values of GFDL CM2.0, for exampal a value 2.985E-07, as this value is in (kg m-2 s-1), i want to convert in mm/month so please help me in this regard. 
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