Date and time computations with GAMAP
NOTE: For all intents and purposes, the terms Greenwich Mean Time (GMT) and Universal Coordinated Time (UTC) are interchangeable. They both refer to the standard time at Greenwich, UK. We shall use GMT in the discussion below.
- 1 Julian Day vs. Astronomical Julian Day
- 2 Astronomical Julian Day routines in IDL and GAMAP
- 3 Computing the day of year from a calendar date (and vice-versa)
- 4 Working with TAU values
- 5 Separating YYYYMMDD into year, month day
- 6 Computing local time at a particular location
- 7 Testing for leap years
Julian Day vs. Astronomical Julian Day
The Julian Day (JD) is used to denote the number of days that have elapsed since the start of a year. For example, the Julian Day number of 2008/01/10 is 10 (since it is the 10th day of 2008).
However, another term that is sometimes confused with the Julian Day is the Astronomical Julian Day (AJD). This is the number of days that have elapsed since 12:00 GMT on January 1, 4713 BC. The Astronomical Julian Day at 00:00 GMT on 2008/01/10 is 2454475.5.
The Astronomical Julian Day is mostly used in astrophysics and space sciences to compute long intervals of time that may span many years (e.g. the period of a variable star or of a comet's orbit, etc.). You can find algorithms for computing AJD in several textbooks, including Practical Astronomy with Your Calculator by Peter Duffett-Smith, Cambridge Univ. Press, 1992.
The chief advantage of using the Astronomical Julian Day is that its computation accounts for the end-of-month, end-of-year, end-of-century, and end-of-millenium transitions. This makes it very useful for computing future or past dates in scripts and computer programs.
For example, if you want to compute what the date will be 1000 days from 2008/01/01, then all you have to do is:
- compute the Astronomical Juliay Day for 2008/01/01
- add 1000
- convert back to a calendar date
We'll return to this example in the next section.
TIP: In order to avoid confusing Julian Day with Astronomical Julian Day, we recommend that you use the terminology "Day of Year" rather than "Julian Day". This will remove all ambiguity.
Astronomical Julian Day routines in IDL and GAMAP
IDL has two Astronomical Julian Day functions
- JULDAY -- converts a year, month, day (, hour, min, sec) to Astronomical Julian Day
- CALDAT -- converts an Astronomical Julian Day back to year, month, day (, hour, min, sec)
NOTE: The year, month, day are mandatory but you may omit the hour, min, sec. If you omit hour, min, sec, then JULDAY will return an long integer value. If you include hour, min, sec, JULDAY wil return a double precision value.
JULDAY and CALDAT are used as follows:
; Compute the Astronomical Julian Day for 2008/01/10 ; Note: the day is required, the hours, mins, seconds are optional IDL> print, julday( 1, 10, 2008, 0, 0, 0 ) 2454475.5 ; Convert the Astronomical Julian Day back to a calendar date IDL> caldat, 2454475.5, y, m, d, h, mi, s IDL> print, y, m, d, h, mi, s 1 10 2008 0 0 0.0000000
To compute the date 1000 days after 2008/01/01 (from the example in the preceding section), you would do the following:
IDL> jd = julday( 1, 1, 2008 ) IDL> jd = jd + 1000 IDL> caldat, jd, y, m, d IDL> print, y, m, d 9 27 2010
We find that gives us the date 2010/09/27. NOTE: This has also accounted for the leap-year-day on Feb 29, 2008.
For your convenience, GAMAP has a function called ADD_DATE which will do this for you in one fell swoop:
IDL> print, add_date( 20080101, 1000 ) 20100927
You can also use ADD_DATE to compute days prior to a given date. Let's compute what the calendar date was 1000 days prior to Jan 1, 2008:
IDL> print, add_date( 20080101, -1000 ) 20050406
Computing the day of year from a calendar date (and vice-versa)
A quick way to compute the day of the year is w/ IDL's JULDAY function:
Day_of_Year = JULDAY( Month, Day, Year ) - JULDAY( 1, 0, Year )
IDL> print, julday( 1, 10, 2008 ) - julday( 1, 0, 2008 ) 10
JULDAY( 1, 0, 2008 ) is another way of saying 12/31/2007. This is used so that you don't have to add one to the above equation. (This is equivalent to JULDAY( 1, 10, 2008 ) - JULDAY( 1, 1, 2008 ) + 1.)
The GAMAP function DAY_OF_YEAR does the above computation for you. You can replace the statement above with:
IDL> print, day_of_year( 1, 10, 2008 ) 10
To compute the calendar date from the day of the year (the inverse operation), you can use IDL's CALDAT function. However, you must also add the Astronomical Julian day for the first of the year, for example:
IDL> caldat, julday( 1, 0, 2008 )+10, y, m, d IDL> print, y, m, d 1 10 2008
But this is made much easier with the GAMAP functions DAY_OF_YEAR and ADD_DATE:
; Convert month/day/year to day of year IDL> doy = day_of_year( 1, 10, 2008 ) IDL> print, doy 10 ; convert day of year back to month/day/year IDL> print, add_date( 20080101, doy-1 ) 20080110
The only thing to remember is you have to subtract 1 from the day of year due to the way that ADD_DATE is written.
NOTE: Starting in GAMAP v2-12 (not released yet), you will be able to specify a single YYYYMMDD argument instead of the month, day, year arguments:
; Convert month/day/year to day of year IDL> doy = day_of_year( 20080110 )
Working with TAU values
Conceptually similar to the Astronomical Julian Day, the TAU value is a monotonically-increasing time index that is used to timestamp output from the GEOS-Chem and GISS models:
- GEOS-Chem: TAU is the count of hours since 00:00 GMT on Jan 1, 1985
- GISS: TAU is the count of hours since 00:00 GMT on Jan 1, 1980
However, unlike the Astronomical Julian Day, TAU is given in hours and not days. Also, the reason for the difference in starting date for TAU is that at the time GEOS-Chem was being created, there existed no meteorological data for it prior to Jan 1, 1985. So 1/1/85 was taken as the start date.
GAMAP ships with 2 functions for working with TAU values: NYMD2TAU and TAU2YYMMDD. They are used as follows:
; Convert a date and time to TAU ; GEOS-Chem style (from 1985) is the default IDL> tau_gc = nymd2tau( 20080101, 030000 ) IDL> print, tau_gc 201603.00 ; Convert TAU back to date and time (GEOS-Chem style) IDL> date = tau2yymmdd( tau_gc, /nformat ) IDL> print, date 20080101 30000
Note that TAU2YYMMDD returns a 2-element vector with the date and time when you use the /NFORMAT keyword. If you omit this keyword then TAU2YYMMDD will return a structure with year, month, day, hour, minute, second tags:
; Convert TAU back to date and time (GEOS-Chem style) ; but this time return as a structure IDL> date = tau2yymmdd( tau_gc ) IDL> help, date, /structure ** Structure <2061898>, 6 tags, length=24, data length=24, refs=1: YEAR LONG Array MONTH LONG Array DAY LONG Array HOUR LONG Array MINUTE LONG Array SECOND LONG Array
NYMD2TAU and TAU2YYMMDD assume GEOS-Chem style (from 1 Jan 1985) as the default. To compute TAU for GISS style (from 1 Jan 1980), you can call these functions with the /GISS keyword as follows:
; Convert date to TAU value (GISS style) IDL> tau_giss = nymd2tau( 20080101, 030000, /giss ) IDL> print, tau_giss 245451.00 ; Convert TAU value back to date (GISS style) IDL> date = tau2yymmdd( tau_giss, /nformat, /giss ) IDL> print, date 20080101 30000
Separating YYYYMMDD into year, month day
GAMAP ships with 2 routines which make it easy to extract (or combine) the year, month, day from (to) a date in YYYYMMDD format: DATE2YMD and YMD2DATE:
; Split 20080101 into separate year, month, day variables IDL> date2ymd, 20080101, y, m, d IDL> print, y, m, d 2008 1 1 ; Combine year, month, day variables into YYYYMMDD format IDL> date = ymd2date( 2008, 1, 1 ) IDL> print, date 20080101
You can also use DATE2YMD and YMD2DATE with times in HHMMSS format:
; Split the time 12:30:45 into separate hour, min, sec variables IDL> date2ymd, 123045, hour, min, sec IDL> print, hour, min, sec 12 30 45 ; Combine 12h, 30min, 45sec into a single time variable IDL> time = ymd2date( 12, 30, 45 ) IDL> print, time 123045
Computing local time at a particular location
Local Time (sometimes called Local Solar Time) is the time at your location defined by the position of the Sun in the sky. It is computed as follows:
Local Time = GMT + ( Longitude / 15 )
- GMT = Greenwich Mean Time
- Longitude is in the range -180...180
The Earth rotates 360 degrees in 24 hours, or 15 degrees in one hour. Therefore, your Local Time is the number of hours that you are east or west of Greenwich.
Your Local Time may differ slightly from your Standard Time Zone Time, depending on where you are located within your Standard Time Zone. In the Continental US, the official Time Zone Times are defined at the Standard Longitude Meridians of 75W (Eastern), 90W (Central), 115W (Mountain), and 130W (Pacific).
- For example, if you live in Boston MA (71.0 W), you would notice that the sun sets at approximately 4:30 PM Eastern on a given day in late December. However, if you live in Cleveland OH (82 W), then you would notice that the sun sets at approximately 5:00 PM for the same date.
- This is because Boston is located to the east of the Eastern Time Zone Meridian (75 W). So the sun will set at Boston earlier than it will at the Standard Time Zone Meridian. Similarly, Cleveland (81.7 W) is located to the west of the Eastern Time Zone Meridian (75 W). Thus, the sun will set in Cleveland later than at the Eastern Time Zone Meridian...and even later than in Boston.
- By convention, both Cleveland and Boston set their clocks to Eastern Time, which is the Local Time at 75W longitude. However, even though Boston and Cleveland may have the same Time Zone Time, their Local Times will differ. This is evidenced by the difference in sunset times at both cities.
NOTE: Local Time does not take Daylight Savings into account.
You can compute the Local Time with GAMAP's LOCALTIME function. You need to specify the Greenwich Mean Time and the longitude of your location.
; Local time at Boston @ 0 GMT ; NOTE: West longitude is negative! IDL> lt_bos = localtime( 0, -71.06 ) IDL> print, lt_bos 19.2627
; Local time at Cleveland @ 0 GMT IDL> lt_cle = localtime( 0, -81.68 ) IDL> print, lt_cle 18.5547
; Difference in local times between Boston & Cleveland ; Boston's Local Time is 0.7 hours (42 min) earlier than Cleveland! IDL> print, lt_bos - lt_cle 0.708000
Note that LOCALTIME returns time in decimal hours. You can convert that to minutes by multiplying the result by 60.
Testing for leap years
In the Gregorian calendar, a Leap Year is defined as any year that is:
- divisible by 4, or
- if a century year, divisible by 400
The following ARE leap years: 1600, ... 1980, 1984, 1988, 1992, 1996, 2000, 2004, 2008, 2012, 2016 ...
The following ARE NOT leap years: 1700, 1800, 1900
You can use GAMAP's ISLEAP function to test if a given year is a leap year or not. ISLEAP returns 1 if a year is a leap year and 0 if not. ISLEAP will accept both scalars and vector arguments:
; Test if 2000 is a leap year IDL> print, isleap( 2000 ) 1 ; Test if 1600, 1700, 1800, 1900, 2000 are leap years ; NOTE: Only century years divisible by 400 are leap years! IDL> print, isleap( [1600, 1700, 1800, 1900, 2000] ) 1 0 0 0 1
--Bmy 14:04, 7 April 2008 (EDT)